As soon as you get that x = log ( whatever), you can stop. There is no reason at this point to continue and bring x into an "ordinary" number. After all, you go into a "roughly equal to" situation, which obviously isn't accurate. It's something that a lot of students also need to be taught, not every solution needs to be an "ordinary" number, like 3.145. Log ( whatever) is also just a fine number. e^(5π) is just a fine number, a fine solution. By continuing the calculations until we reach an "ordinary" number, first of all it doesn't really make any sense because you cannot calculate it by yourself and you need a calculator, and secondly, it gives the impression to students that log (whatever) is just not really a number, it's weird and just creates more chaos and a stereotype that this is hard to understand and should be avoided, when in reality, it's not. You can always try to "imagine" if there can be simplifications in some cases, but if there isn't any more simplification to be done, you should just leave it be.
Exactly (pun intended ...). Those equal signs between approximate solutions only add to the confusion. Also, if you approximate anyways in the end, you can as well use numerical bisection or Newton's method from the start.
The problem doesn't state that x has to be real; it could be complex. So it's wrong to eliminate the negative root of the quadratic equation, (1-√5)/2. That's because, in general, there is a complex x such that a^x is negative for positive real a. So there is a second solution based on (3/2)^x = (1-√5)/2 which comes out as: x = [log((1-√5)/2)]/log(3/2). Or it's equally true if you use the natural log: x = [ln((1-√5)/2)]/ln(3/2) That may be enough to satisfy some people. However, most calculators cannot evaluate log((1-√5)/2) or ln((1-√5)/2) because the log of a negative number is complex (and in fact has an infinite set of values). Ideally, we want to find computable expressions for the real and imaginary parts of the solution. Fortunately, there is a formula for the log of a negative number, expressed as real and imaginary parts, which can be derived from Euler's identity e^iπ + 1 = 0. This is just a special case of the general identity e^ikπ + 1 = 0, which is true for all odd integers k. Rearranging gives e^ikπ = -1, then taking the natural log of both sides, we get ln(-1) = ikπ, for all odd integers k, which is an infinite set of values. Now we can use this to derive ln(-a) for any positive real a, knowing that ln(xy) = ln(x) + ln(y): ln(-a) = ln(-1.a) = ln(-1) + ln a = ikπ + ln a, for all odd integers k. In our case, -a = (1-√5)/2, and a = (√5-1)/2, so we have: ln((1-√5)/2) = ikπ + ln((√5-1)/2), for all odd integers k. Putting this into our solution for x, we have: x = [ikπ + ln((√5-1)/2)]/ln(3/2), for all odd integers k. So we have found an infinite set of complex solutions to the original equation, in addition to the single real solution in the video.
If we want to get numbers from a calculator for this, then we can take the simplest case of k = 1, and separate out the real and imaginary parts: x = ln((√5-1)/2)/ln(3/2) + iπ/ln(3/2) x = -1.1868 + 7.7481i (to 4 places) but take note that we can also add or subtract any multiple of 2iπ/ln(3/2) which comes out as 15.4962i, to get other solutions.
@@darthcaedus8813 Na strednej škole sa komplexnými číslami zaoberajú veľmi okrajovo a nie je v ich matematickej výbave aby to dokázali riešiť. Riešenie v obore reálnych čísel je správne a príklad je veľmi efektne vypočítaný. Pobavil som sa. Ďakujem.
Fairly easy for an Olympiad. Write exponents of 3 and 2 (the numbers are special for a reason) That yields to a quadratic-only equation: a^2-ab=b^2 You naturally want to reduce to a quadratic with one variable, so it's only natural to divide by b^2 so everything can be written with a/b ratio. You get the golden equation: X^2=1+X where X = (3/2)^x X > 0 so the negative solution is rejected, leaving only x=phi taking natural logs: x= ln(phi)/ln(3/2)
I don't know about about olympiad problems. I am a retired hobbyist. This was refreshing, but went on too long, so long in fact, I thought it must be wrong answer, but like OP, I checked and the answer is good. But too many steps and with lots of digits/numbers/operations. not expect this in math competition. Smart people don't use calculator too much. Once I see golden ratio, I want to work it out from the quadratic but resist making so complicated. Sure enough, the answer is there and yields same numeric answer. Thank you for opportunity to polish my old math tools. I learn something from your video and subscribe your channel.
Thanks for this exercise, this is actually pretty neat :-) (That's right, I'm a lay man with some historical education.) Just like many others, I first noticed that x=1 is a tad too small and x=2 is already way too many. Then I used the "solver" available in MS Excel to get the result down to about 4 decimal places. Then I tried cracking the symbolic math, before actually watching the vid - and I took a slightly different approach, obviously using a combination of the same elementary laws. After staring cross-eyed at the numbers for a few minutes, I figured "okay, what if I break down the numbers into primes and try to rehash them around a little: 9^x - 6^x = 4^x 3^x * 3^x - 3^x * 2^x = 2^x * 2^x What happens if I try to cancel out some of those 3^x and 2^x, by dividing the whole equation by them? And I ended up with 1 - 2^x/3^x = (2^x/3^x)^2 (2^x/3^x)^2 + 2^x/3^x -1 = 0 The substitution was obvious: u^2 + u - 1 = 0 The solutions are: u{1,2} = (-1+- sqrt(5) ) / 2 And, reverting the substitution, u{1,2} = (2/3)^x Only the positive solution of "u" can be logarithmized: x = log[2/3]((sqrt(5)-1)/2) Only then, I took a look at the vid. Initially I found the author's approach slightly redundant and acrobatic, but I got a little alert at around 6:30, where she goes: u = (1+sqrt(5)) / 2 Oh wait. Is she missing a leading minus? I let her calculate all the way to the end, coming up with the same deciamal value that I have previously arrived at. But... that quadratic solution for "u" clearly was different than mine? It took me about 10 minutes to find a second difference, that compensates for the one described above. She has settled on log base = 3/2, whereas using my approach, I ended up with 2/3. That, combined with the flipped sign to the "1" in the solution for "u", means that our two results are equal! x == log[2/3]((sqrt(5)-1)/2) == log[3/2]((sqrt(5)+1)/2) This is *RAD* :-D
@@GrumpyMcFrog yes I'm pretty sure. 9^x = (3*3)^x = 3^x * 3^x. I cannot find a formula for this particular case, but I have found (a^n * a^m) = a^(m+n) . If I start from there, and apply this to the example you're asking about, a^x * a^x = a^(2x) = (a^2)^x If you don't believe me, try substituting some example exponents and run that through a calculator. For x=2 or x=3, I can do the calculation in my head.
In many cases, the solution of a math problem, as exactly as it is, can only be written in terms of certain mathematical symbolic language other than a decimal number. In such cases we must use mathematical symbols, not decimal numbers. For example, the solution of the equation x2 =2 is x=√2. If one writes x=1.414 then that is not the solution but an approximation of the solution. One may use calculator to get x=1.414213562 or use the virtual calculator in a computer to get 32 digits after the decimal point. But then they are all approximations of the solution. Here let us use mathematical symbols to write down the solution exactly as it is: X=(log (1+√5)/2)/log〖3/2〗 Sorry, youtube does not provide good math format to accept many math symbols. In the above x2 =2 actually means x square equals 2.
I just plugged in "1" and observed the two sides. Then I used "2" and could see I went too far. Next, I tried 1.2 and knew I was close. Then I tried 1.19 and called it close enough. Less than a minute this iterative way. And it would work even easier with a graphing calculator. And it doesn't depend on having the problem numbers made special for solving by long hand.
@@VYDEO In fairness to Newton, his interpolation method is more involved than just bigger/smaller number. He adjusts the guessrd for x based on the ratio between the deltas of the Ys and Xs for consecutive guesses, until he reaches equilibrium (or close enough). I assume you knew that but simplified for brevity. Just mentioning so people don't think Newton was blindly throwing darts at a wall (or an apple).
She solved the exact value with pen and paper. To me that is the true solution, if purpose is to learn math. In engineering we would need the value of x (down to N decimals accuracy), but that is why calculators exist.
@@m1k1a1 An engineer would also probably use an iterative method to find that solution rather than go through this algebraic process. It is also much faster here, particularly with access to modern devices, and completely acceptable mathematically. As a mathematician I greatly appreciate the various creators making maths content but I do think problems that show advanced maths being less efficient than a simple approach are generally detrimental, particularly when the video title is potentially misleading as to what is required!
@@zelandakhniteblade5436 Also when realizing, that nowadays you are rarely asked to possess teh ability to do the maths and rather that you need to use software like Mathcad to solve, simply, diffferentiate, integrate, sketch...
7:57 You don't need to go through all that. By definition of logarithm, if a^x=b then x=log(a)b. And the final answer should be in a logarithm form because it's an irrational number. If you want to answer 1.1868 that's fine but you need to use approximate symbol not equal.
A very good problem for demonstrating in a practical manner the use of many of the laws of exponents to solve a problem. Gaining a good understanding of how to apply the laws of exponents is essential for practicing engineers. Well done!
@@PhillipAlcock Imagine, what even more better career u would have had if u would use laws of exponents.... Maybe u would even been beating Elon Musk right now..
As engineers you dont even have to mangle into symbolic operations :) only keep it well counted into 2 or 3 decimal points :) - even in this video lady pointed the answer to be 1,1868 - on math olympic question you are not supposed to answer like that :)
Beautifully done. I did it by trying 1 then 2. Found it close to 1. Next step would be by succesive approx. I like your true method and the steps better. Thank you
I'm study Engineering and at the same time I teach math to high school students for University exam. I like this questions. It's amazing. I thought about what I could do before following the solution but I didn't think it would take this long. But let's admit it. It could have been solved much faster. For example I think it would be simpler to use Ln. I think you tried to make it easy and understandable. Thanks.
the way you wrote it was totally understandable to me and I did not even notice the mistakes ....so it is perfectly fine. I hope somebody will answer your questions which was not about grammar :)
While this is clearly above anything I would be able to solve in my school days, it's not hard to understand that someone with a little bit sharper mind can. At least I did recognize some formulas that I haven't seen in at least 17 years, but I have completely forgot how log works. I obviously don't use maths beyond percentages and such in my daily life, so forgetting anything but the most simple stuff is expected.
Not sure why you'd *expect* to forget stuff. I still remember stuff from high school, which was 39+ years ago. I can still quote Avogodro's number (6.02 * 10^23), or the ideal gas law (PV=NRT). I've never used calculus for anything other than fun, yet I still can.
naaa! if you were a simple janitor and to have to clean a patio with a broom, of course you know the weight of the broom, you know the friction of the broom against the floor, plus some wind, you already know some physics' laws, only because you are a janitor people can say you are knowledgeable, but no, you are some people that use some Nweton's laws @@fastaskirby2414
After reading the comments here, it's obvious that most people on TH-cam had a pretty easy time with this problem. I on the other hand had a different experience. I followed along nicely for about five minutes, then after thinking here comes the solution four of five times, by the eight minute mark I was quite upset, thinking what's going on here? The thing is growing with new appendages, like a low budget horror film.
Thank you for this video. It is a very nice problem, and the solution is elegant. You have inspired me to write up some personal notes (just for fun) about solving other equations of that structure. Best wishes!
I calculated x from my mind, with a calculation error, result is 1.87, then corrected it to a value as 1.187, so close without steps :) Thanks for the solution!.. Great..
The only sophisticated step is finding the common base to permit the variable substitution. The original coefficients are rigged to permit this. So you have to see a step that will only work when the problem is specially rigged, as opposed to something you could do in the general case.
"This could not be equated" - it is not precise. Actually it could (we do have complex numbers and at the start of video there was no restrictions mentioned that solution must be real) log(-1) = πi
@Alex9ilov, наверняка требуется решить уравнение аналитически, а не графически. Тем более что графически в данном случае получится лишь приближённое решение.
@@rinatsamerhanov5371у нее как раз точное. Просто зря она за калькулятор взялась. Все, что после 9:26 - лишнее. Ну и про второй корень, тот что комплексное число, который она заигнорила.
Watching this brought back memories from almost 60 years ago when I stopped taking math classes. Before watching it, I wrote a one-line shell loop that allowed me to find the same answer you did in a few minutes by brute force successive approximation. That said, I appreciate the approach you used.
@@colindant3410 A loop that can be written in one line. I wrote a first try using integers 0..9, then line-edited it to home in on the smallest result. Here is the final line: for x in 1.1860 1.1861 1.1862 1.1863 1.1864 1.1865 1.1866 1.1867 1.1868 1.1869 ; do print $x $(($((pow(9,x)-pow(6,x)))-$((pow(4,x))))) ; done Note that this is ksh (Korn shell) syntax, but other shells work similarly. There are much more intelligent ways to do this, but this was quick and easy and worked.
that sounds like my way. i havent even watched this yet or worked it out but instantly i think, ok, we need higher numbers so maybe try 1.5 9x1.5 is 13.5. 6 x 1.5 is 9. 13.5 - 9 equals 4.5. i slightly overdid it therefore 1.3 is as near as dam it haha Thats sort of how I do any 'instant maths'
If we're cheating and using computers though, Wolfram Alpha can do the whole thing in about 5 seconds and shows the real and complex solutions to 15 decimal places. I'm sure there is an even better tool out there, but as far as speed goes, that's a heck of a useful tool that anyone can use if they know how to type on a keyboard (and, perhaps, don't know how to make Excel do anything)
I would plug in numbers for x until the two sides matched. 9-6 = 3 < 4, so you know x=1 but not 2 where 81-64 = 17 > 16, x is between the values of 1 and 2. So you would have to solve for x to get 4 dec places. Nice problem.
Beautiful example. Also the kind of problem AI will solve easily. My math is rusty. I recognize the moves, but couldn't do them nowadays. I worked in an AI lab in an electrical engineering department. I was taking differential equations at the time and asked the engineering graduate students if they remembered all the math. They said yes because they'd used it so much. However, this is the algebraic way of thinking. I think it terms of geometry. Years later I asked a grad student in math if he could solve Bertrand's paradox. He started writing equations and disappeared down the rabbit hole. I'd "solved" it in a couple minutes using a kind of geometric statistical logic. I say "solved" because though I'd come up with a correct solution, I didn't realize there were two other correct solutions. The reason it's a paradox. In my job, I was asked to solve a path-planning problem that was done algebraically, but took a long time given the processing speed of the day. I found a geometric rule-based solution that cut the time substantially. Basically the rules reduced the number of calculations. When we published it, we got a nasty review from a guy at Stanford who had a PhD student that had come up with something similar. I'd come from a programming job at a mutual fund and had studied art history in college. I'd been at the lab for 30 days when I figured out "my" method. Once I called my wife's uncle for help with a differential equation. He hated differential equations. Cookbooks solutions he said. He was a number theory guy, one of the 100 best mathematicians of the 20th century. At the Institute for Advanced Studies in Princeton, he'd met Einstein waiting at the bus stop. This reminds me of C. P. Snow's book The Two Cultures. Science and the humanities need to be able to talk to each other. Your video is a great reminder that a good education includes everyone having experience of... the method, the struggle, the beauty and magic of mathematics. Everyone doesn't need to master it, but neither should they dismiss it. And smart kids shouldn't be judged based on choices made as kids. I was a C algebra student and an A geometry student. I hated math homework and didn't do it. Geometry I could figure out during the tests. Algebra, the kind of beautiful manipulation you demonstrate, no way. I normally learned the algebra weeks after the test. In 12th grade I took calculus, trig, and linear algebra at a time where very few high-schools were teaching calculus. Two hours of class time a day. After several months of that my algebra was really good. I scored in the 99th percentile on the SAT math part and almost 800 on the math achievement test. The whole reason I was in the advanced math section was because the head of the math department knew I was a smart kid who didn't work. She fought for me and didn't give up. I have great respect for mathematics and have often been the most mathematically competent among groups of PhDs. But nothing like you. Inspiring video. Thanks.
I have wolfram alpha on my phone, and it solves it saying assuming x is real and will show all the intermediate steps if you want. It also shows a graphical solution by plotting 4^x vs. 9^x - 6^x and using the intersection of the curves.
@@zorro5651 Nice. I think I would have benefited from this kind of software. I'm more interested in understanding the meaning than in how to solve things. I took differential equations early in the PC era. Half the classes used software on a Mac that gave you a class of solutions via graphs. I got the traditional way. One day the TA taught, but you didn't have to come. I was one of the few who did. He taught on how differential equations explained tuning a radio to a broadcast frequency. Best math class I ever had.
Step 1: We'll rearrange the equation to have all the terms on one side: 9^x - 6^x - 4^x = 0 Step 2: Since the bases (9, 6, and 4) are different, we can't immediately simplify the equation. To proceed, let's look for a substitution that might help. A common approach is to factor out the smallest term, which in this case is 4^x: 4^x * ( (9/4)^x - (6/4)^x - 1 ) = 0 Step 3: Now, we have a product of two terms that must be equal to zero. This means either the first term (4^x) is zero or the second term ((9/4)^x - (6/4)^x - 1) is zero. First, let's consider the second term: ( (9/4)^x - (6/4)^x - 1 ) = 0 Step 4: We can simplify the second term further: ( (9/4)^x - (3/2)^x - 1 ) = 0 Unfortunately, there is no straightforward algebraic way to solve this equation. However, we can still find approximate solutions using numerical methods or graphing. Let's try to approximate a solution for x using a numerical method. Step 5: For numerical approximation, you can use tools like graphing calculators or software like Wolfram Alpha or numerical methods such as the Newton-Raphson method. Alternatively, you can use trial and error to find approximate solutions. For simplicity, let's use trial and error. If we try x = 2, we get: ( (9/4)^2 - (3/2)^2 - 1 ) ≈ (2.25 - 2.25 - 1) ≈ -0.5 As this is not equal to zero, we need to try a different value of x. Let's try x = 3: ( (9/4)^3 - (3/2)^3 - 1 ) ≈ (5.06 - 3.375 - 1) ≈ 0.685 This is also not equal to zero. You can continue trying different values of x using trial and error or use numerical methods to get a more accurate approximation. Keep in mind that the exact solution to this equation might not be expressible in simple algebraic terms.
@@ThePrimeMinisterOfTheBlock Hello Mr. Prime Mister. My wife's uncle was one of the 100 best mathematicians of the 20th century. I once asked him for help in differential equations. He hated differential equations. Said they were Cookbook math. Anything that's a cookbook or involves trial and error - or numerical approximations - is perfect for AI. It didn't "solve" Go, Chess, or Shogi, but it revolutionized the first by playing millions of games and is better than any human in all three. In mathematical theorem proving, the problem was of a different kind. The "proof" would involve thousands of steps on problems that only a few people in the world understood. And to check the "proof" involved a lot of work. Also potentially flawed. Sorry, Mr Prime Minister, it's turtles already down.
Microsoft Bing solved the problem in 5 seconds as below: 9^x-6^x=4^x 9^x-6^x=4^x To solve the equation 9^x - 6^x = 4^x for x, we can use logarithms. Here is one way to solve it: Rewrite the equation as 9^x = 6^x + 4^x. Take the logarithm of both sides of the equation. You can use any base for the logarithm, but it’s usually easiest to use the natural logarithm (ln). Apply the logarithm rules to simplify the equation. For example, ln(a + b) = ln(a) + ln(1 + b/a). Solve for x. Using this method, we get: 9^x = 6^x + 4^x ln(9^x) = ln(6^x + 4^x) x ln(9) = ln(6^x + 4^x) x = ln(6^x + 4^x) / ln(9) x ≈ 1.1868
Remember those days, 65 years ago. Calculus tests with solutions requiring dozens of steps. With pencil and paper. I usually made an error in one of the steps. Back then, slide rules were prohibited. Later calculators.
The question was taken from an exam, thus the student is meant to demonstrate their understanding of the techniques taught in the class. Taking a shortcut is not an option!
@@sambelld1 The question was taken from the Olympiad exam. And I know exactly how this olympiad thing works. They just needed an approach that's all. Though the solution is correct but not in efficient way.
Апроксимацию можно было начать гораздо раньше. Такие задачи не дают на олимпиадах. На олимпиадах дают задачи по поиску аналитического решения, которое является точным.
Я вообще в первый момент не понял, что происходит, когда она начала десятичные дроби лепить))) Привык, что в математике требуется точный ответ))) Да и что-то она где-то напуделяла, у меня после 1,8 цифры малость поболее вышли, даже если так считать...
Silly to do all of those partial approximations at the end. Use your calculator in a more grown-up way, and go from the exact form of the solution to the approximate in just ONE step.
@@Calvaryscout yes, true. Her difference was smaller than mine. I got a final result of 3.7~4.4. Hers was in the order of 0.0001. More acceptable, right?
9^x - 6^x = 4^x Step-by-step solution using the Newton-Raphson method: Step 1: Choose an initial guess for x. Let's start with x = 1. Step 2: Calculate the function value and its derivative at the initial guess. f(x) = 9^x - 6^x - 4^x f'(x) = (9^x * ln(9)) - (6^x * ln(6)) - (4^x * ln(4)) At x = 1: f(1) ≈ 9^1 - 6^1 - 4^1 = 9 - 6 - 4 = -1 f'(1) ≈ (9^1 * ln(9)) - (6^1 * ln(6)) - (4^1 * ln(4)) Step 3: Use the Newton-Raphson formula to update the value of x. x_new = x - f(x) / f'(x) At x = 1: x_new = 1 - (-1) / f'(1) Step 4: Repeat steps 2 and 3 until the value of x converges to a desired level of accuracy. Continue this iteration process until the value of x converges to a desired level of accuracy. After several iterations, the approximate value of x is found to be: x ≈ 0.774489 This value satisfies the equation 9^x - 6^x = 4^x approximately.
A quick check of the math (e(x) for e^x and l(x) for ln(x)) you gave shows your x doesn't satisfy the original equation: e(l(9)*0.774489) - e(l(6)*0.774489) = 1.4777865440646662 e(l(4)*0.774489) = 2.9260978203064426 These two results are not close enough.. compare the given answer of l((1 + sqrt(5))/2) / l(3/2) = 1.18681439028098171754: e(l(9)*1.18681439028098171754) - e(l(6)*1.18681439028098171754) = 5.18243020056306326302 e(l(4)*1.18681439028098171754) = 5.18243020056306326297 These two are sufficiently close to believe the presented answer is approximately correct. calculating fractional exponents with a calculator button seems to be a thing people take as a given rather than working out the math with the identity of x^y =e^(y * ln(x))
Проверка на приближенных значениях нелепый и бессмысленный шаг, поскольку все преобразования в решении тождественны. В чем была ценность аналитического решения, если в конце мы берём калькулятор и считает!?
It's funny, but I solved such equations in the 11th, last grade in Ukraine =) At a school with in-depth study of math. Now I tried to solve it myself 5 years after graduation, and got the same answer
I would recommend more care with the notation. When equating something to a fraction, the "=" sign must be centered with respect to the fraction line. In general, all signs between fractions should be centered with respecto to te fraction lines. Regarding the solution of the quadratic equation, I think it's better (from the mathematical point of view) to complete the squares instead of applying a memorized formula
I haven't seen the video, and I guess the way ist similar: 9^x - 6^x = 4^x --> ln9/ln2 = 3.16 ; ln6/ln2 = 2.58 ; ln4/ln2 = 2 (2^x)^3.16 - (2^x)^2.58 = (2^x)^2 substitution: z = 2^x
z^3.16 - z^2.58 = z² /z² z^1.16 - z^0.58 = 1 first exponent is like double the second: (z^0.58)² - z^0.58 = 1 substituion: a = (z^0.58) a² - a - 1 = 0 pq-formula: a = 1.62 substituion back: z = a^(1/0.58) z = 1.62^(1/0.58) z = 2.28 x = ln(z)/ln(2) x = ln(2.28)/ln(2) x = 1.186814390280982
It isn't this that they'll use, it is the skills of problem solving, which they will use every day of their lives, and people who don't understand this need help.
@@MrGrumbleguts There is no skill of problem solving here. Nothing about the solution presented transfers to more general cases that you are likely to encounter in the real world. Like an earlier comment said, the equation is rigged for this solution. It only works in this special scenario. Try with others, like use 5 instead of 4, and you'll be stuck. This is a moot academic exercise with no practical bearing. I always despised Olympiads for these kinds of questions but my school kept sending me anyway. *eyeroll*
And there are many engineering applications to problems where you recast it to a problem that has an analytical solution, so you don’t need a fancy computer. The answer will almost always be good enough because you don’t initially have enough information to formulate accurate equations anyway.
Maths can be fun, but I will give you the summary of the important skills of my maths degree, which I apply daily.. Imagine a cake Applied maths - the edges/boundaries/icing decide what happens. And always reality check your solution. Statistics - knowing what the cake is made of decides the data. But if we cover it with icing. Seriously, understand the limitations of your data. Pure maths - if we can create a recipe for cake for all possible cakes we can eat the cake. Seriously, being able to formulate the problem is half the battle. different parts of ut the most important thing you learn in maths is to be able to state problems
I do not have a problem with math but with math bullshit. I like when someone shows something that we can apply to real life and understand how that formula, equation or whatever fits in our lives. Other than that is the same o, same o, "mathsturbation". No offense to you but it is what it is. That's why many of us like physics better, it could be hard but all fits in daily actions. Things like: a car riding at x speed per hour gets from point A to point B. That is useful.
idk about chinese olympiad, but if it was mcq based then I would have first assumed the exponent to be 1 only, and 9-6 is 3, so the answer would be close to 1.1 or 1.2 only anyways, claps for the patience you had throughout the explanation
These are taught in school level not in the college's especially in engineering colleges and polytechnics guys don't know basic math ....but they are owing the ............banks or bank wives ???
@@georgen9755 Electrical engineering is extremely math intensive. I certainly learned everything needed to solve this problem easily. The math required for many courses in electrical engineering is much harder than these basics. These are all things I used and got proficient at while studying EE: Trigonometry, Geometry, Analytic Geometry, Calculus, Differential Equations, Partial Differential Equations, LaPlace Transforms, Lagrangian multipliers, Vectors, Vector Calculus, Tensors, Tensor Calculus, Matrixes and Matrix Algebra, Linear Programming, Non-Linear Programming, Algorithms, Logarithms, Statistics ... My engineering college was extremely heavy on math. In fact I would argue there was way to much math and way too little practical applications when I went to a top 15 engineering college in the 1980's.
Simple solution - clicked play and found out! ;) Not done maths to any level for *cough ahem* a few decades or so and that used rules that I did not know existed, gracefully done!
Could also have applied the Secant Approximation to solve ((9^x - 6^x)/(4^x)) - 1 = 0, with initial tries of x = 1 & x = 2. In today's world where numerical methods and brute force computing triumph over classical analysis, speed of computation is what counts. But your analysis is beautiful nonetheless. Also, as @oussamajaber3512 observed earlier, once you solved the quadratic for 1.5^x you could have used either natural logs or logs to base 10 to immediately solve the answer.
If you are programming something where you need exactly this calculation method several times, for example, it makes sense to optimize the algorithm even with today's mechanical computing power. Sometimes it is even necessary. The way is then to decompose the problem, as in the solution to a problem like this. In other words, what used to be intelligent still is today.
If you don't know the fundamentally based mathematical process,.... then you don't know theory. If you don't know the theory, you can't solve all or a least a substantially wider range of problems. The point of the video is not speed, but to show how to properly get the answer,.. not the answer itself. Imagine using trial and error to go to space.
You could have used a spreadsheet, plot 2 functions, the left side and the right side of the equations and see where they intersect. I did that in 10 seconds and got the same result.
this is also a question to ask the software engineer or any engineer, so that you can ask it and expect the candidate to not be able to answer it, and then, hire the young or the good looking guy or girl, and write in the report "although he or she cannot solve it, he or she shows good progress". For the old person, just write in the report: "Cannot solve in allocated time. Not recommending a hire. I don't see a strong signal."
Perfectly clear but I don't have the patience to write down every step when I see where it is going. That's why I found myself fastforwarding a lot. But this is the better/correct way and avoids mistakes.
You will learn to think analytically.. 👍😀 (e.g. if I do this and that, then it will result in this and that.. OK, the future is not predictable, but since everything on earth and on life is a matter of probability, everything depends how many past experiences you had stored (own experiences, books, good films, stories from friends, etc.) in your brain to extrapolate a possible outcome.. And the sum of all this is called INTELLIGENCE!) 👍😀
This is kindergarten stuff! I gave this to my 2-yo daughter and she solved it in seconds. Admittedly, her answer was different to the one shown here, but the squiggles looked nice.
Regardless of the number of people that say you should have used a calculator earlier...i think this is a very good demonstration of how to solve these types of questions
Nice flow of calculation demo! @ around 11:40 you wrote log2 as .301 in the numerator and .307 in the denominator. Could that account for some of the inaccuracy in your final answer?
No. Inaccuracy appeared as soon as she started using numbers and is inevitable when you write numbers instead of formulas when dealing with irrational numbers. As for the questioned values, I think this is just a writing issue, maybe decoding someone's (own?) notes on this, she also writes 0.4221 instead of 0.4771, but the outcome of 0.1761 is ok.
Nice job. However, assuming we have limited ourselves to Real solutions, it’s all fine until 7:58 at which point the explanation becomes unnecessarily long and complicated. We can take the logarithm of both sides in ANY base, for instance: x ln[3/2] = ln[1+sqr5)/2] (natural logarithm) x Log[3/2] = Log[1+sqr5)/2] (base 10 logarithm) This means that if we’re looking for a numerical approximation, we simply calculate this in whatever base we want: x = log[1+sqr5)/2] / log[3/2] ~ 1.18681 and we’re done.
It is obvious that this equation does not have a solution because X can not be a whole number. On a technicality the answer of 1.1868..... is not the solution because it wil never perfectly solve this equation as it goes to infinity, same like 33.333333 multiplied by 3 will never equal 100.
Actually the answer is wrong. The problem was formulated with exact equality. The answer is the valid solution of quadratic equation. Going further requires including in formulation of a problem required precision otherwise validity of solution can be argued. Math starts with discipline.
In my high school and college days, I had several math classes. Not once did any ever teach about logs. I got lost when you started talking about them. I can see they are useful, but how you got there, I'm still here standing in the dust.
Easy peasy, especially of you can do radicals and logs in your head! -_-' Besides that, superb problem. I did do a few substitutions in my head like 1, -1, 2, -2, 1/2, -1/2 and I realized the solution is not an easy to come by number. And that's it. lol
Chat gpt: Sure, I'd be happy to help you solve the equation! Let's work through it step by step: Start with the given equation: 9^x - 6^x = 4^x Notice that 4^x can be written as (2^x)^2, so the equation becomes: 9^x - 6^x = (2^x)^2 Now, let's rewrite 9^x and 6^x as powers of 2 using the fact that 9 = 3^2 and 6 = 2 * 3: 9^x = (3^2)^x = 3^(2x) 6^x = (2 * 3)^x = 2^x * 3^x Substitute these values back into the equation: 3^(2x) - 2^x * 3^x = (2^x)^2 Factor out 3^x from the left-hand side of the equation: 3^x * (3^x - 2^x) = (2^x)^2 Now, notice that (2^x)^2 can be rewritten as 2^(2x): 3^x * (3^x - 2^x) = 2^(2x) Divide both sides by 3^x: 3^x - 2^x = 2^(2x) / 3^x Now, let's simplify the right-hand side by expressing 2^(2x) as (2^x)^2: 3^x - 2^x = (2^x)^2 / 3^x Substitute y = 2^x into the equation: 3^x - y = y^2 / 3^x Rearrange the equation: 3^(2x) - 3^x * y - y^2 = 0 Now, this equation is quadratic in terms of y. We can solve it using the quadratic formula: y = [3^x ± sqrt((3^x)^2 - 4 * 3^(2x) * (-1))] / 2 Simplify further: y = [3^x ± sqrt(9^x + 4 * 3^x)] / 2 Now, remember that y = 2^x, so we have: 2^x = [3^x ± sqrt(9^x + 4 * 3^x)] / 2 We can simplify this further by multiplying both sides by 2: 2^(x+1) = 3^x ± sqrt(9^x + 4 * 3^x) At this point, you can solve for x using numerical methods or algebraic manipulation. It's a challenging equation to solve directly, but you can use numerical methods or graphical techniques to approximate the value of x.
I am not a mathematician, not even that crazy about solving formulas, but the original formula was stating "equals", not "approximates", therefore this is not really a solution. If you are not setting any parameters for deviation, you can say, that x=2. The result will be way off, but you can say, that it is "approximately" the same with difference in the order of magnitude. :)
Durch einfaches probieren kommt man schnell auf eine Zahl zwischen 1 und 2. 9-6=4 oder 81-36=16. Dabei wird auch deutlich, dass der Wert eher bei 1 als bei 2 liegt.
11:54 Error: the 2nd log(2) you wrote as 0.307, instead of 0.301. Then at 12:00, you subtracted wrong. As written, it would be 0.1701. Instead, you wrote the correct value, as if your previous mistake weren't there.
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As soon as you get that x = log ( whatever), you can stop. There is no reason at this point to continue and bring x into an "ordinary" number. After all, you go into a "roughly equal to" situation, which obviously isn't accurate. It's something that a lot of students also need to be taught, not every solution needs to be an "ordinary" number, like 3.145. Log ( whatever) is also just a fine number. e^(5π) is just a fine number, a fine solution. By continuing the calculations until we reach an "ordinary" number, first of all it doesn't really make any sense because you cannot calculate it by yourself and you need a calculator, and secondly, it gives the impression to students that log (whatever) is just not really a number, it's weird and just creates more chaos and a stereotype that this is hard to understand and should be avoided, when in reality, it's not. You can always try to "imagine" if there can be simplifications in some cases, but if there isn't any more simplification to be done, you should just leave it be.
Exactly (pun intended ...). Those equal signs between approximate solutions only add to the confusion.
Also, if you approximate anyways in the end, you can as well use numerical bisection or Newton's method from the start.
The problem doesn't state that x has to be real; it could be complex. So it's wrong to eliminate the negative root of the quadratic equation, (1-√5)/2. That's because, in general, there is a complex x such that a^x is negative for positive real a. So there is a second solution based on (3/2)^x = (1-√5)/2 which comes out as:
x = [log((1-√5)/2)]/log(3/2). Or it's equally true if you use the natural log:
x = [ln((1-√5)/2)]/ln(3/2)
That may be enough to satisfy some people. However, most calculators cannot evaluate log((1-√5)/2) or ln((1-√5)/2) because the log of a negative number is complex (and in fact has an infinite set of values). Ideally, we want to find computable expressions for the real and imaginary parts of the solution.
Fortunately, there is a formula for the log of a negative number, expressed as real and imaginary parts, which can be derived from Euler's identity e^iπ + 1 = 0.
This is just a special case of the general identity e^ikπ + 1 = 0, which is true for all odd integers k. Rearranging gives
e^ikπ = -1, then taking the natural log of both sides, we get
ln(-1) = ikπ, for all odd integers k, which is an infinite set of values.
Now we can use this to derive ln(-a) for any positive real a, knowing that ln(xy) = ln(x) + ln(y):
ln(-a) = ln(-1.a) = ln(-1) + ln a = ikπ + ln a, for all odd integers k.
In our case, -a = (1-√5)/2, and a = (√5-1)/2, so we have:
ln((1-√5)/2) = ikπ + ln((√5-1)/2), for all odd integers k.
Putting this into our solution for x, we have:
x = [ikπ + ln((√5-1)/2)]/ln(3/2), for all odd integers k.
So we have found an infinite set of complex solutions to the original equation, in addition to the single real solution in the video.
If we want to get numbers from a calculator for this, then we can take the simplest case of k = 1, and separate out the real and imaginary parts:
x = ln((√5-1)/2)/ln(3/2) + iπ/ln(3/2)
x = -1.1868 + 7.7481i (to 4 places)
but take note that we can also add or subtract any multiple of 2iπ/ln(3/2) which comes out as 15.4962i, to get other solutions.
i guess on the level of high school, presumtion that x belongs to real is ok
@@darthcaedus8813 Na strednej škole sa komplexnými číslami zaoberajú veľmi okrajovo a nie je v ich matematickej výbave aby to dokázali riešiť. Riešenie v obore reálnych čísel je správne a príklad je veľmi efektne vypočítaný. Pobavil som sa. Ďakujem.
But if x can be complex can it also be quaternion?
@@milanganyicz2957 This is an olympiad question and not an ordinary high school quiz show so complex number is a fair game to explore.
Fairly easy for an Olympiad.
Write exponents of 3 and 2 (the numbers are special for a reason)
That yields to a quadratic-only equation: a^2-ab=b^2
You naturally want to reduce to a quadratic with one variable, so it's only natural to divide by b^2 so everything can be written with a/b ratio.
You get the golden equation: X^2=1+X where X = (3/2)^x
X > 0 so the negative solution is rejected, leaving only x=phi
taking natural logs:
x= ln(phi)/ln(3/2)
I don't know about about olympiad problems. I am a retired hobbyist. This was refreshing, but went on too long, so long in fact, I thought it must be wrong answer, but like OP, I checked and the answer is good. But too many steps and with lots of digits/numbers/operations. not expect this in math competition. Smart people don't use calculator too much. Once I see golden ratio, I want to work it out from the quadratic but resist making so complicated. Sure enough, the answer is there and yields same numeric answer. Thank you for opportunity to polish my old math tools. I learn something from your video and subscribe your channel.
It is very easy for an Olympiad problem; in fact it is not of Olympiad standard at all.
@@johnnath4137 it is a fairly standard problem for a 3rd-4th year of high school in Italy.
@@proftorricelli40 sure sure, and yet I never meet any Italians who are any good at maths
You mean approximate instead of equal to.
Why use the complicated log instead of using the natural log? Applying ln would become:
(3/2)^x = ((1+sqr5)/2)
ln (3/2)^x = ln ((1+sqr5)/2)
x . ln (3/2)= ln ((1+sqr5)/2)
x= ln ((1+sqr5)/2) / ln (3/2)
x= 1.1868
not everyone knows the value by heart
That's very cool too
it's even more "stupid" when later in the video we transform the complicated log into the natural log
Indeed, and any log give the same result here anyway, regardless if it is ln, log_10 or log_(3/2) since log_n (x) = ln(x) / ln(n).
ln is just easier because e^(lnx) = x, and ln (e^x) = x
Thanks for this exercise, this is actually pretty neat :-) (That's right, I'm a lay man with some historical education.)
Just like many others, I first noticed that x=1 is a tad too small and x=2 is already way too many.
Then I used the "solver" available in MS Excel to get the result down to about 4 decimal places.
Then I tried cracking the symbolic math, before actually watching the vid - and I took a slightly different approach, obviously using a combination of the same elementary laws. After staring cross-eyed at the numbers for a few minutes, I figured "okay, what if I break down the numbers into primes and try to rehash them around a little:
9^x - 6^x = 4^x
3^x * 3^x - 3^x * 2^x = 2^x * 2^x
What happens if I try to cancel out some of those 3^x and 2^x, by dividing the whole equation by them? And I ended up with
1 - 2^x/3^x = (2^x/3^x)^2
(2^x/3^x)^2 + 2^x/3^x -1 = 0
The substitution was obvious:
u^2 + u - 1 = 0
The solutions are:
u{1,2} = (-1+- sqrt(5) ) / 2
And, reverting the substitution,
u{1,2} = (2/3)^x
Only the positive solution of "u" can be logarithmized:
x = log[2/3]((sqrt(5)-1)/2)
Only then, I took a look at the vid.
Initially I found the author's approach slightly redundant and acrobatic, but I got a little alert at around 6:30, where she goes:
u = (1+sqrt(5)) / 2
Oh wait. Is she missing a leading minus?
I let her calculate all the way to the end, coming up with the same deciamal value that I have previously arrived at.
But... that quadratic solution for "u" clearly was different than mine?
It took me about 10 minutes to find a second difference, that compensates for the one described above.
She has settled on log base = 3/2, whereas using my approach, I ended up with 2/3.
That, combined with the flipped sign to the "1" in the solution for "u", means that our two results are equal!
x == log[2/3]((sqrt(5)-1)/2) == log[3/2]((sqrt(5)+1)/2)
This is *RAD* :-D
after reading all this my brain is twisting!
😅🤣🤣
Are you sure 9^x = 3^x*3^x ?
@@GrumpyMcFrog yes I'm pretty sure. 9^x = (3*3)^x = 3^x * 3^x. I cannot find a formula for this particular case, but I have found (a^n * a^m) = a^(m+n) . If I start from there, and apply this to the example you're asking about, a^x * a^x = a^(2x) = (a^2)^x
If you don't believe me, try substituting some example exponents and run that through a calculator. For x=2 or x=3, I can do the calculation in my head.
That's why I prefer anything else, except Math. I could solve it myself, but it took me about 20 minutes !
This is the first time I have seen someone approaching this problem your way.
Beautiful.
In many cases, the solution of a math problem, as exactly as it is, can only be written in terms of certain mathematical symbolic language other than a decimal number. In such cases we must use mathematical symbols, not decimal numbers. For example, the solution of the equation x2 =2 is x=√2.
If one writes x=1.414 then that is not the solution but an approximation of the solution. One may use calculator to get x=1.414213562 or use the virtual calculator in a computer to get 32 digits after the decimal point. But then they are all approximations of the solution. Here let us use mathematical symbols to write down the solution exactly as it is:
X=(log (1+√5)/2)/log〖3/2〗
Sorry, youtube does not provide good math format to accept many math symbols. In the above x2 =2 actually means x square equals 2.
I just plugged in "1" and observed the two sides. Then I used "2" and could see I went too far. Next, I tried 1.2 and knew I was close. Then I tried 1.19 and called it close enough. Less than a minute this iterative way. And it would work even easier with a graphing calculator. And it doesn't depend on having the problem numbers made special for solving by long hand.
That repetitive approach would work, constantly narrowing down the potential answer.
My thought exactly. I just graphed the formula for y = 9^x - 6^x - 4^x, then zoomed in to where y = 0 and got my answer.
That is exactly what is known as Newton's interpolation method and then there are many more interpolation and extrapolation methods 😂
@@VYDEO In fairness to Newton, his interpolation method is more involved than just bigger/smaller number. He adjusts the guessrd for x based on the ratio between the deltas of the Ys and Xs for consecutive guesses, until he reaches equilibrium (or close enough).
I assume you knew that but simplified for brevity. Just mentioning so people don't think Newton was blindly throwing darts at a wall (or an apple).
@@ptrinch Obviously I knew it was oversimplification. 🙏
Also, the solution from the set of complex numbers:
x = (iπ-ln((1+√5)/2))/ln(3/2)
I understand that you're transforming this into the complex plane. but how did π appear?
@@gregparrott Due to the connection of the complex exponent with trigonometric functions.
In particular, Euler once showed that: e^(iπ)=-1
For someone of my generation "You should be able to solve this" means without the use of a calculator.
For your (grand)parents' generation that would have meant "solve without pen and paper" 🤪
You can't be pre-log tables as well though.
She solved the exact value with pen and paper. To me that is the true solution, if purpose is to learn math. In engineering we would need the value of x (down to N decimals accuracy), but that is why calculators exist.
@@m1k1a1 An engineer would also probably use an iterative method to find that solution rather than go through this algebraic process. It is also much faster here, particularly with access to modern devices, and completely acceptable mathematically. As a mathematician I greatly appreciate the various creators making maths content but I do think problems that show advanced maths being less efficient than a simple approach are generally detrimental, particularly when the video title is potentially misleading as to what is required!
@@zelandakhniteblade5436 Also when realizing, that nowadays you are rarely asked to possess teh ability to do the maths and rather that you need to use software like Mathcad to solve, simply, diffferentiate, integrate, sketch...
7:57 You don't need to go through all that. By definition of logarithm, if a^x=b then x=log(a)b. And the final answer should be in a logarithm form because it's an irrational number. If you want to answer 1.1868 that's fine but you need to use approximate symbol not equal.
Exactly my sentiment. This is (supposed to be) mathematics, not engineering (where close enough is sufficient)
A very good problem for demonstrating in a practical manner the use of many of the laws of exponents to solve a problem. Gaining a good understanding of how to apply the laws of exponents is essential for practicing engineers. Well done!
Funny, I had a career as an engineer and I can’t remember having to use laws of exponents….
@@PhillipAlcock Good for you.
@@PhillipAlcock Imagine, what even more better career u would have had if u would use laws of exponents.... Maybe u would even been beating Elon Musk right now..
As engineers you dont even have to mangle into symbolic operations :) only keep it well counted into 2 or 3 decimal points :) - even in this video lady pointed the answer to be 1,1868 - on math olympic question you are not supposed to answer like that :)
@@дядяМразь-н8ж
Bet Elon never had to use the laws of exponents either. Either that or he hired someone who could if needed.
Um... if you are going to use a calculator at the end why not just plug it all into the calculator as soon as you had the log expression.
Exactly my thoughts
I have absolutely no idea what I just saw .. but I watched it all for some reason. Also, very pleasant handwriting.
Beautifully done. I did it by trying 1 then 2. Found it close to 1. Next step would be by succesive approx.
I like your true method and the steps better. Thank you
I'm study Engineering and at the same time I teach math to high school students for University exam. I like this questions. It's amazing. I thought about what I could do before following the solution but I didn't think it would take this long. But let's admit it. It could have been solved much faster. For example I think it would be simpler to use Ln. I think you tried to make it easy and understandable. Thanks.
Good thing you're not teaching English
@@CAustin582 .Why you say that bro .d My mother language is not engllish. I just try
the way you wrote it was totally understandable to me and I did not even notice the mistakes ....so it is perfectly fine. I hope somebody will answer your questions which was not about grammar :)
I think so too!
I was thinking the same thing. Why take the lo(n)g way when a log or ln would suffice from the very beginning.
While this is clearly above anything I would be able to solve in my school days, it's not hard to understand that someone with a little bit sharper mind can. At least I did recognize some formulas that I haven't seen in at least 17 years, but I have completely forgot how log works. I obviously don't use maths beyond percentages and such in my daily life, so forgetting anything but the most simple stuff is expected.
Knowing the existence and knowing how to apply is completely different thing. This is what separate smart people from knowledgeable people.
Not sure why you'd *expect* to forget stuff. I still remember stuff from high school, which was 39+ years ago. I can still quote Avogodro's number (6.02 * 10^23), or the ideal gas law (PV=NRT). I've never used calculus for anything other than fun, yet I still can.
naaa! if you were a simple janitor and to have to clean a patio with a broom, of course you know the weight of the broom, you know the friction of the broom against the floor, plus some wind, you already know some physics' laws, only because you are a janitor people can say you are knowledgeable, but no, you are some people that use some Nweton's laws @@fastaskirby2414
After reading the comments here, it's obvious that most people on TH-cam had a pretty easy time with this problem. I on the other hand had a different experience. I followed along nicely for about five minutes, then after thinking here comes the solution four of five times, by the eight minute mark I was quite upset, thinking what's going on here? The thing is growing with new appendages, like a low budget horror film.
i love your horror movie reference. I felt the same!
By chance, was the film Stanley Kubrick's 'The Shining'?
Yes. If I had been solving this in an exam, at the point it started to become 'horrific' I would have thought I had gone wrong and abandoned it.
Same.
That's one hell of a way to describe the problem!! :-)
Thank you for this video. It is a very nice problem, and the solution is elegant. You have inspired me to write up some personal notes (just for fun) about solving other equations of that structure. Best wishes!
I calculated x from my mind, with a calculation error, result is 1.87, then corrected it to a value as 1.187, so close without steps :) Thanks for the solution!.. Great..
Oh heck, just put enough gas in it to get to Mars and back and cancel the astronaut's insurance policies.
Such a relaxing and interesting video. You are clever, and you have the most wonderful, beautiful voice. ❤
Thank you Dear Superman❤️
Thank you please go forward giving people knowledge 💖🥺🥺👍👍👍
@Superman. Ditto. She has a beautiful voice.
The only sophisticated step is finding the common base to permit the variable substitution. The original coefficients are rigged to permit this. So you have to see a step that will only work when the problem is specially rigged, as opposed to something you could do in the general case.
I hate these kinds of problems, where the solution is a gimmick. Finding the solution numerically (using graphs) is just as valid then.
All Olympiad problem are like that: rigged to be easy, once you see the trick.
I really enjoyed watching you refresh me on logs and exponents.
GREAT WORK !!!
"This could not be equated" - it is not precise. Actually it could (we do have complex numbers and at the start of video there was no restrictions mentioned that solution must be real)
log(-1) = πi
Строим два графика функций: y1=9x-6x и y2=4x. Точка пересечения этих двух функций y1 и y2 - и есть решение задачи.
@Alex9ilov, наверняка требуется решить уравнение аналитически, а не графически. Тем более что графически в данном случае получится лишь приближённое решение.
@@ВасилийСорокин-ю1с а у нее тоже не точное решение
ахахха
@@rinatsamerhanov5371Почему не точное? x=log_3/2[((1+sqrt(5))/2].
@@rinatsamerhanov5371у нее как раз точное. Просто зря она за калькулятор взялась. Все, что после 9:26 - лишнее. Ну и про второй корень, тот что комплексное число, который она заигнорила.
Simple. Divide both parts by 2^x or 3^x, and do the substitution, getting rid of the degree.
I love math but I have not paid attention during my highschool days so I'm terrible at it but seeing problem solved is so satisfying though.
Excellent!! Great problem. Please add more problems like this problem to your channel. Thanks.
Watching this brought back memories from almost 60 years ago when I stopped taking math classes. Before watching it, I wrote a one-line shell loop that allowed me to find the same answer you did in a few minutes by brute force successive approximation. That said, I appreciate the approach you used.
A quick head calculation tells you the answer lies slightly greater than one and less than two.
What is a "one line shell loop"?
@@colindant3410 A loop that can be written in one line. I wrote a first try using integers 0..9, then line-edited it to home in on the smallest result. Here is the final line:
for x in 1.1860 1.1861 1.1862 1.1863 1.1864 1.1865 1.1866 1.1867 1.1868 1.1869 ; do print $x $(($((pow(9,x)-pow(6,x)))-$((pow(4,x))))) ; done
Note that this is ksh (Korn shell) syntax, but other shells work similarly. There are much more intelligent ways to do this, but this was quick and easy and worked.
@@gshenaut Thanks. This reminds me of For...Next loops and Do loops.
that sounds like my way. i havent even watched this yet or worked it out but instantly i think, ok, we need higher numbers so maybe try 1.5 9x1.5 is 13.5. 6 x 1.5 is 9. 13.5 - 9 equals 4.5. i slightly overdid it therefore 1.3 is as near as dam it haha Thats sort of how I do any 'instant maths'
Very good problem. Thanks. If we (3/2)^x=1+2,2364(value for squareroot of 5)/2. And then, take a log on either side, we can get 1,18.
Got it to 6 decimal places using Excel in about 10 minutes (iteration). In practical terms much easier for most people than 'solving' mathematically.
MS Excel is a pretty dang handy thing. 😊
If we're cheating and using computers though, Wolfram Alpha can do the whole thing in about 5 seconds and shows the real and complex solutions to 15 decimal places. I'm sure there is an even better tool out there, but as far as speed goes, that's a heck of a useful tool that anyone can use if they know how to type on a keyboard (and, perhaps, don't know how to make Excel do anything)
I can use TH-cam and see the answer in 5 seconds
Hypercalc Pro on my Smartphone solves it in no time.
Everybody who is interested in math should give it a try.
your voice is so calm, helps even doing chinese math lol
A nice reminder of why knowing logarithmic properties is so important
Contrary....never had to use it ever in my life.
@@lwh7301 Then you've never lived 😜 YOLO!
@@CalvinEastwood Never used it once in med school, and never used it once in 40 years after.
I would plug in numbers for x until the two sides matched. 9-6 = 3 < 4, so you know x=1 but not 2 where 81-64 = 17 > 16, x is between the values of 1 and 2. So you would have to solve for x to get 4 dec places. Nice problem.
Beautiful example. Also the kind of problem AI will solve easily. My math is rusty. I recognize the moves, but couldn't do them nowadays. I worked in an AI lab in an electrical engineering department. I was taking differential equations at the time and asked the engineering graduate students if they remembered all the math. They said yes because they'd used it so much. However, this is the algebraic way of thinking. I think it terms of geometry. Years later I asked a grad student in math if he could solve Bertrand's paradox. He started writing equations and disappeared down the rabbit hole. I'd "solved" it in a couple minutes using a kind of geometric statistical logic. I say "solved" because though I'd come up with a correct solution, I didn't realize there were two other correct solutions. The reason it's a paradox.
In my job, I was asked to solve a path-planning problem that was done algebraically, but took a long time given the processing speed of the day. I found a geometric rule-based solution that cut the time substantially. Basically the rules reduced the number of calculations. When we published it, we got a nasty review from a guy at Stanford who had a PhD student that had come up with something similar. I'd come from a programming job at a mutual fund and had studied art history in college. I'd been at the lab for 30 days when I figured out "my" method. Once I called my wife's uncle for help with a differential equation. He hated differential equations. Cookbooks solutions he said. He was a number theory guy, one of the 100 best mathematicians of the 20th century. At the Institute for Advanced Studies in Princeton, he'd met Einstein waiting at the bus stop.
This reminds me of C. P. Snow's book The Two Cultures. Science and the humanities need to be able to talk to each other. Your video is a great reminder that a good education includes everyone having experience of... the method, the struggle, the beauty and magic of mathematics. Everyone doesn't need to master it, but neither should they dismiss it. And smart kids shouldn't be judged based on choices made as kids.
I was a C algebra student and an A geometry student. I hated math homework and didn't do it. Geometry I could figure out during the tests. Algebra, the kind of beautiful manipulation you demonstrate, no way. I normally learned the algebra weeks after the test. In 12th grade I took calculus, trig, and linear algebra at a time where very few high-schools were teaching calculus. Two hours of class time a day. After several months of that my algebra was really good. I scored in the 99th percentile on the SAT math part and almost 800 on the math achievement test. The whole reason I was in the advanced math section was because the head of the math department knew I was a smart kid who didn't work. She fought for me and didn't give up. I have great respect for mathematics and have often been the most mathematically competent among groups of PhDs. But nothing like you. Inspiring video. Thanks.
I have wolfram alpha on my phone, and it solves it saying assuming x is real and will show all the intermediate steps if you want. It also shows a graphical solution by plotting 4^x vs. 9^x - 6^x and using the intersection of the curves.
@@zorro5651 Nice. I think I would have benefited from this kind of software. I'm more interested in understanding the meaning than in how to solve things. I took differential equations early in the PC era. Half the classes used software on a Mac that gave you a class of solutions via graphs. I got the traditional way. One day the TA taught, but you didn't have to come. I was one of the few who did. He taught on how differential equations explained tuning a radio to a broadcast frequency. Best math class I ever had.
AI will solve easily hey? Let's find out
Step 1: We'll rearrange the equation to have all the terms on one side:
9^x - 6^x - 4^x = 0
Step 2: Since the bases (9, 6, and 4) are different, we can't immediately simplify the equation. To proceed, let's look for a substitution that might help. A common approach is to factor out the smallest term, which in this case is 4^x:
4^x * ( (9/4)^x - (6/4)^x - 1 ) = 0
Step 3: Now, we have a product of two terms that must be equal to zero. This means either the first term (4^x) is zero or the second term ((9/4)^x - (6/4)^x - 1) is zero.
First, let's consider the second term:
( (9/4)^x - (6/4)^x - 1 ) = 0
Step 4: We can simplify the second term further:
( (9/4)^x - (3/2)^x - 1 ) = 0
Unfortunately, there is no straightforward algebraic way to solve this equation. However, we can still find approximate solutions using numerical methods or graphing. Let's try to approximate a solution for x using a numerical method.
Step 5: For numerical approximation, you can use tools like graphing calculators or software like Wolfram Alpha or numerical methods such as the Newton-Raphson method. Alternatively, you can use trial and error to find approximate solutions.
For simplicity, let's use trial and error. If we try x = 2, we get:
( (9/4)^2 - (3/2)^2 - 1 ) ≈ (2.25 - 2.25 - 1) ≈ -0.5
As this is not equal to zero, we need to try a different value of x. Let's try x = 3:
( (9/4)^3 - (3/2)^3 - 1 ) ≈ (5.06 - 3.375 - 1) ≈ 0.685
This is also not equal to zero. You can continue trying different values of x using trial and error or use numerical methods to get a more accurate approximation.
Keep in mind that the exact solution to this equation might not be expressible in simple algebraic terms.
@@ThePrimeMinisterOfTheBlock Hello Mr. Prime Mister. My wife's uncle was one of the 100 best mathematicians of the 20th century. I once asked him for help in differential equations. He hated differential equations. Said they were Cookbook math. Anything that's a cookbook or involves trial and error - or numerical approximations - is perfect for AI. It didn't "solve" Go, Chess, or Shogi, but it revolutionized the first by playing millions of games and is better than any human in all three. In mathematical theorem proving, the problem was of a different kind. The "proof" would involve thousands of steps on problems that only a few people in the world understood. And to check the "proof" involved a lot of work. Also potentially flawed. Sorry, Mr Prime Minister, it's turtles already down.
Комментарий в пять слов - просто ради поддержки канала.
И спасибо автору❤ за доставленное удовольствие :-)
Microsoft Bing solved the problem in 5 seconds as below:
9^x-6^x=4^x
9^x-6^x=4^x
To solve the equation 9^x - 6^x = 4^x for x, we can use logarithms. Here is one way to solve it:
Rewrite the equation as 9^x = 6^x + 4^x.
Take the logarithm of both sides of the equation. You can use any base for the logarithm, but it’s usually easiest to use the natural logarithm (ln).
Apply the logarithm rules to simplify the equation. For example, ln(a + b) = ln(a) + ln(1 + b/a).
Solve for x.
Using this method, we get:
9^x = 6^x + 4^x ln(9^x) = ln(6^x + 4^x) x ln(9) = ln(6^x + 4^x) x = ln(6^x + 4^x) / ln(9)
x ≈ 1.1868
Thank you for your logic explanation, I've been trying to solve this equation since 1973 (my final year at school)
Remember those days, 65 years ago. Calculus tests with solutions requiring dozens of steps. With pencil and paper. I usually made an error in one of the steps. Back then, slide rules were prohibited. Later calculators.
This is the worse solution I've ever seen. If you are going to use a calculator eventually then you should use it in the previous page.
Exactly. That solution confuse me. Why do yo have to do all of that to explain a so dumb in math like me all of that😅😅
The question was taken from an exam, thus the student is meant to demonstrate their understanding of the techniques taught in the class. Taking a shortcut is not an option!
@@sambelld1 The question was taken from the Olympiad exam. And I know exactly how this olympiad thing works. They just needed an approach that's all. Though the solution is correct but not in efficient way.
Ending with the affirmation this is the best channel ever.. makes me regret watching.
Апроксимацию можно было начать гораздо раньше. Такие задачи не дают на олимпиадах.
На олимпиадах дают задачи по поиску аналитического решения, которое является точным.
@@beaver333 у них принято до приблизительного численного ответа доводить? Молодцы, отличная практика.
Я вообще в первый момент не понял, что происходит, когда она начала десятичные дроби лепить))) Привык, что в математике требуется точный ответ))) Да и что-то она где-то напуделяла, у меня после 1,8 цифры малость поболее вышли, даже если так считать...
@@beaver333, писать -1 вместо i нормально. Если это не требуется часто, зачем плодить лишний формализм? Мозгов от этого не прибавится.
Silly to do all of those partial approximations at the end. Use your calculator in a more grown-up way, and go from the exact form of the solution to the approximate in just ONE step.
I managed to solve it differently but got an error at the order of decimals😅
Great exercise! Thanks for sharing
in her near final solution she uses log 2 as BOTH .301 and .307 so there is an error in her numbers.....
@@Calvaryscout yes, true. Her difference was smaller than mine. I got a final result of 3.7~4.4. Hers was in the order of 0.0001. More acceptable, right?
9^x - 6^x = 4^x
Step-by-step solution using the Newton-Raphson method:
Step 1: Choose an initial guess for x. Let's start with x = 1.
Step 2: Calculate the function value and its derivative at the initial guess.
f(x) = 9^x - 6^x - 4^x
f'(x) = (9^x * ln(9)) - (6^x * ln(6)) - (4^x * ln(4))
At x = 1:
f(1) ≈ 9^1 - 6^1 - 4^1 = 9 - 6 - 4 = -1
f'(1) ≈ (9^1 * ln(9)) - (6^1 * ln(6)) - (4^1 * ln(4))
Step 3: Use the Newton-Raphson formula to update the value of x.
x_new = x - f(x) / f'(x)
At x = 1:
x_new = 1 - (-1) / f'(1)
Step 4: Repeat steps 2 and 3 until the value of x converges to a desired level of accuracy.
Continue this iteration process until the value of x converges to a desired level of accuracy. After several iterations, the approximate value of x is found to be:
x ≈ 0.774489
This value satisfies the equation 9^x - 6^x = 4^x approximately.
A quick check of the math (e(x) for e^x and l(x) for ln(x)) you gave shows your x doesn't satisfy the original equation:
e(l(9)*0.774489) - e(l(6)*0.774489) = 1.4777865440646662
e(l(4)*0.774489) = 2.9260978203064426
These two results are not close enough.. compare the given answer of l((1 + sqrt(5))/2) / l(3/2) = 1.18681439028098171754:
e(l(9)*1.18681439028098171754) - e(l(6)*1.18681439028098171754) = 5.18243020056306326302
e(l(4)*1.18681439028098171754) = 5.18243020056306326297
These two are sufficiently close to believe the presented answer is approximately correct.
calculating fractional exponents with a calculator button seems to be a thing people take as a given rather than working out the math with the identity of x^y =e^(y * ln(x))
Проверка на приближенных значениях нелепый и бессмысленный шаг, поскольку все преобразования в решении тождественны. В чем была ценность аналитического решения, если в конце мы берём калькулятор и считает!?
It's funny, but I solved such equations in the 11th, last grade in Ukraine =)
At a school with in-depth study of math. Now I tried to solve it myself 5 years after graduation, and got the same answer
I would recommend more care with the notation. When equating something to a fraction, the "=" sign must be centered with respect to the fraction line. In general, all signs between fractions should be centered with respecto to te fraction lines. Regarding the solution of the quadratic equation, I think it's better (from the mathematical point of view) to complete the squares instead of applying a memorized formula
Концовка убила всю элегантность решения.
Чувак, все же знают значения логарифмов. Ну ты чё?
@@backtracks963 а зачем они нужны, эти значения? Никто же не пишет в решении"6.28" вместо "2π", вот и тут так же.
We Chinese no need the results of 1.1868.but the results of log ab is the final answer
I haven't seen the video,
and I guess the way ist similar:
9^x - 6^x = 4^x
--> ln9/ln2 = 3.16 ; ln6/ln2 = 2.58 ; ln4/ln2 = 2
(2^x)^3.16 - (2^x)^2.58 = (2^x)^2
substitution:
z = 2^x
z^3.16 - z^2.58 = z² /z²
z^1.16 - z^0.58 = 1
first exponent is like double the second:
(z^0.58)² - z^0.58 = 1
substituion:
a = (z^0.58)
a² - a - 1 = 0
pq-formula:
a = 1.62
substituion back:
z = a^(1/0.58)
z = 1.62^(1/0.58)
z = 2.28
x = ln(z)/ln(2)
x = ln(2.28)/ln(2)
x = 1.186814390280982
Fibonacci auxiliary eq.
An+2 = An+1 + An
t^2 = t + 1
Golden ratio t = (1+root5)/2
t = (3/2)^x = (1+root5)/2
log . . .
I eventually understand why kids at school would say I'm never gonna need this.
It isn't this that they'll use, it is the skills of problem solving, which they will use every day of their lives, and people who don't understand this need help.
@@MrGrumbleguts There is no skill of problem solving here. Nothing about the solution presented transfers to more general cases that you are likely to encounter in the real world. Like an earlier comment said, the equation is rigged for this solution. It only works in this special scenario. Try with others, like use 5 instead of 4, and you'll be stuck. This is a moot academic exercise with no practical bearing. I always despised Olympiads for these kinds of questions but my school kept sending me anyway. *eyeroll*
@@CotopaxiAH1968 Wow, what an incredibly stupid and narrow view.
@@CotopaxiAH1968 Olympiad questions are not meant for practical applications. They are meant for people who enjoy mathematics
And there are many engineering applications to problems where you recast it to a problem that has an analytical solution, so you don’t need a fancy computer. The answer will almost always be good enough because you don’t initially have enough information to formulate accurate equations anyway.
Write a python program to evaluate this. Testing the all X values between 1 and 2. Utilise the interval method. Then hone in on the solution.
Now I remember why I hated math in college. Nobody can think math is fun after seeing this.
Maths can be fun, but I will give you the summary of the important skills of my maths degree, which I apply daily..
Imagine a cake
Applied maths - the edges/boundaries/icing decide what happens. And always reality check your solution.
Statistics - knowing what the cake is made of decides the data. But if we cover it with icing.
Seriously, understand the limitations of your data.
Pure maths - if we can create a recipe for cake for all possible cakes we can eat the cake. Seriously, being able to formulate the problem is half the battle.
different parts of ut the most important thing you learn in maths is to be able to state problems
You're right. To have fun out of this you shouldn't write every detailed step, but I guess that was not the point of the video :(
I do not have a problem with math but with math bullshit. I like when someone shows something that we can apply to real life and understand how that formula, equation or whatever fits in our lives. Other than that is the same o, same o, "mathsturbation". No offense to you but it is what it is. That's why many of us like physics better, it could be hard but all fits in daily actions. Things like: a car riding at x speed per hour gets from point A to point B. That is useful.
The girl explains in too much detail!
idk about chinese olympiad, but if it was mcq based then I would have first assumed the exponent to be 1 only, and 9-6 is 3, so the answer would be close to 1.1 or 1.2 only
anyways, claps for the patience you had throughout the explanation
Thank you very much for teaching us Maths!
These are taught in school level not in the college's especially in engineering colleges and polytechnics guys don't know basic math ....but they are owing the ............banks or bank wives ???
@@georgen9755 Electrical engineering is extremely math intensive. I certainly learned everything needed to solve this problem easily. The math required for many courses in electrical engineering is much harder than these basics. These are all things I used and got proficient at while studying EE: Trigonometry, Geometry, Analytic Geometry, Calculus, Differential Equations, Partial Differential Equations, LaPlace Transforms, Lagrangian multipliers, Vectors, Vector Calculus, Tensors, Tensor Calculus, Matrixes and Matrix Algebra, Linear Programming, Non-Linear Programming, Algorithms, Logarithms, Statistics ... My engineering college was extremely heavy on math. In fact I would argue there was way to much math and way too little practical applications when I went to a top 15 engineering college in the 1980's.
Simple solution - clicked play and found out! ;) Not done maths to any level for *cough ahem* a few decades or so and that used rules that I did not know existed, gracefully done!
Could also have applied the Secant Approximation to solve ((9^x - 6^x)/(4^x)) - 1 = 0, with initial tries of x = 1 & x = 2. In today's world where numerical methods and brute force computing triumph over classical analysis, speed of computation is what counts. But your analysis is beautiful nonetheless. Also, as @oussamajaber3512 observed earlier, once you solved the quadratic for 1.5^x you could have used either natural logs or logs to base 10 to immediately solve the answer.
If you are programming something where you need exactly this calculation method several times, for example, it makes sense to optimize the algorithm even with today's mechanical computing power. Sometimes it is even necessary. The way is then to decompose the problem, as in the solution to a problem like this.
In other words, what used to be intelligent still is today.
If you don't know the fundamentally based mathematical process,.... then you don't know theory.
If you don't know the theory, you can't solve all or a least a substantially wider range of problems.
The point of the video is not speed, but to show how to properly get the answer,.. not the answer itself.
Imagine using trial and error to go to space.
@@jcgoogle1808 Check out the solution, which I also commented here, before you get the wrong idea of what you think I might have meant.
@@janwendlandt3126
My post was in response to dawnlightening's comment on numerical methods triumphing over fundamental.
@@jcgoogle1808 I see. All right.
You could have used a spreadsheet, plot 2 functions, the left side and the right side of the equations and see where they intersect. I did that in 10 seconds and got the same result.
Great. Tkank You
👍🏻🥰
Class of a teacher.. simple and better
this is also a question to ask the software engineer or any engineer, so that you can ask it and expect the candidate to not be able to answer it, and then, hire the young or the good looking guy or girl, and write in the report "although he or she cannot solve it, he or she shows good progress". For the old person, just write in the report: "Cannot solve in allocated time. Not recommending a hire. I don't see a strong signal."
Perfectly clear but I don't have the patience to write down every step when I see where it is going. That's why I found myself fastforwarding a lot. But this is the better/correct way and avoids mistakes.
How can I use this to save my marriage?
Bcos if I can solve this I can solve any problems in life
You will learn to think analytically.. 👍😀
(e.g. if I do this and that, then it will result in this and that.. OK, the future is not predictable, but since everything on earth and on life is a matter of probability, everything depends how many past experiences you had stored (own experiences, books, good films, stories from friends, etc.) in your brain to extrapolate a possible outcome.. And the sum of all this is called INTELLIGENCE!) 👍😀
@@ft7339that’s a very Good answer!
(1 + sq. root 5)/2 is the golden ratio number (Fi)! Thus, the answer could be written as x = log (Fi)/(log3 - log2)
U took me to my 🎒📚 days😊
Thank you for the clear, step-by-step explanation.
you don't have to take log to the base 3/2. At this step you can use log base 10 and get everything from the calculator
This was so much fun. I needed to follow along all the way through to 4:05 (substitution) before I could finally take over and figure it out.
This is kindergarten stuff! I gave this to my 2-yo daughter and she solved it in seconds. Admittedly, her answer was different to the one shown here, but the squiggles looked nice.
Nice video. Thanks 👏
you lost me at logs. the solution beyond quadratic eq is horrible.
Regardless of the number of people that say you should have used a calculator earlier...i think this is a very good demonstration of how to solve these types of questions
Nice flow of calculation demo!
@ around 11:40 you wrote log2 as .301 in the numerator and .307 in the denominator. Could that account for some of the inaccuracy in your final answer?
No. Inaccuracy appeared as soon as she started using numbers and is inevitable when you write numbers instead of formulas when dealing with irrational numbers.
As for the questioned values, I think this is just a writing issue, maybe decoding someone's (own?) notes on this, she also writes 0.4221 instead of 0.4771, but the outcome of 0.1761 is ok.
Nice job. However, assuming we have limited ourselves to Real solutions, it’s all fine until 7:58 at which point the explanation becomes unnecessarily long and complicated. We can take the logarithm of both sides in ANY base, for instance:
x ln[3/2] = ln[1+sqr5)/2] (natural logarithm)
x Log[3/2] = Log[1+sqr5)/2] (base 10 logarithm)
This means that if we’re looking for a numerical approximation, we simply calculate this in whatever base we want:
x = log[1+sqr5)/2] / log[3/2] ~ 1.18681
and we’re done.
Nice work! However, u forget a simple x value , it is x = 0 .
that doesnt work
1 - 1 = 1?
yikes.
No kid, equation will become 1-1=1 if put x=0 , go learn maths! 😂
@@MikeBrawlStars240 hahaha
I like how she has an answer on 9:22, but refuses to use it and plays back and forth until 11:55
It is obvious that this equation does not have a solution because X can not be a whole number. On a technicality the answer of 1.1868..... is not the solution because it wil never perfectly solve this equation as it goes to infinity, same like 33.333333 multiplied by 3 will never equal 100.
You are very brilliant! Congratulations!
very elegant... enjoyed the tour
I just glanced through the video. Looks like a fun problem.
Actually the answer is wrong. The problem was formulated with exact equality. The answer is the valid solution of quadratic equation. Going further requires including in formulation of a problem required precision otherwise validity of solution can be argued. Math starts with discipline.
In my high school and college days, I had several math classes. Not once did any ever teach about logs. I got lost when you started talking about them. I can see they are useful, but how you got there, I'm still here standing in the dust.
Easy peasy, especially of you can do radicals and logs in your head! -_-'
Besides that, superb problem. I did do a few substitutions in my head like 1, -1, 2, -2, 1/2, -1/2 and I realized the solution is not an easy to come by number. And that's it. lol
Excellent vid by the legendary LK "Life is Kind" Logic
Intelligence is about how quickly you can apply and correlate the rules AND, of course, practice regularly!
the simpler the question appears, the harder the solution is.
Please, place the equal symbol at the same line that the división line. Thanks, of course
Chat gpt:
Sure, I'd be happy to help you solve the equation! Let's work through it step by step:
Start with the given equation: 9^x - 6^x = 4^x
Notice that 4^x can be written as (2^x)^2, so the equation becomes: 9^x - 6^x = (2^x)^2
Now, let's rewrite 9^x and 6^x as powers of 2 using the fact that 9 = 3^2 and 6 = 2 * 3:
9^x = (3^2)^x = 3^(2x)
6^x = (2 * 3)^x = 2^x * 3^x
Substitute these values back into the equation:
3^(2x) - 2^x * 3^x = (2^x)^2
Factor out 3^x from the left-hand side of the equation:
3^x * (3^x - 2^x) = (2^x)^2
Now, notice that (2^x)^2 can be rewritten as 2^(2x):
3^x * (3^x - 2^x) = 2^(2x)
Divide both sides by 3^x:
3^x - 2^x = 2^(2x) / 3^x
Now, let's simplify the right-hand side by expressing 2^(2x) as (2^x)^2:
3^x - 2^x = (2^x)^2 / 3^x
Substitute y = 2^x into the equation:
3^x - y = y^2 / 3^x
Rearrange the equation:
3^(2x) - 3^x * y - y^2 = 0
Now, this equation is quadratic in terms of y. We can solve it using the quadratic formula:
y = [3^x ± sqrt((3^x)^2 - 4 * 3^(2x) * (-1))] / 2
Simplify further:
y = [3^x ± sqrt(9^x + 4 * 3^x)] / 2
Now, remember that y = 2^x, so we have:
2^x = [3^x ± sqrt(9^x + 4 * 3^x)] / 2
We can simplify this further by multiplying both sides by 2:
2^(x+1) = 3^x ± sqrt(9^x + 4 * 3^x)
At this point, you can solve for x using numerical methods or algebraic manipulation. It's a challenging equation to solve directly, but you can use numerical methods or graphical techniques to approximate the value of x.
Hi
When changing base from 10 to 3/2 you should apply the correct rules
I am not a mathematician, not even that crazy about solving formulas, but the original formula was stating "equals", not "approximates", therefore this is not really a solution. If you are not setting any parameters for deviation, you can say, that x=2. The result will be way off, but you can say, that it is "approximately" the same with difference in the order of magnitude. :)
An olympiad is not a competition. It is a bracket of four years, a measurement of time, not a contest.
Durch einfaches probieren kommt man schnell auf eine Zahl zwischen 1 und 2. 9-6=4 oder 81-36=16. Dabei wird auch deutlich, dass der Wert eher bei 1 als bei 2 liegt.
11:54 Error: the 2nd log(2) you wrote as 0.307, instead of 0.301. Then at 12:00, you subtracted wrong. As written, it would be 0.1701. Instead, you wrote the correct value, as if your previous mistake weren't there.
Solved it in two minutes.
School was so long ago but a little FOR LOOP in BASIC got me the answer.
the answer is ln(phi)/ln(1.5), where phi is the golden ratio (i.e. solution to u^2 = u+1)