I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚
I worked it out a bit different. My solution was simply determine what of 3 exponent would get me a number greater than 65 that would be an odd number (3^4). I then subtracted that 65 from that number (81) and I got 16 which is 2^4. In other words you can rewrite the equation in this instance as 3^m - 2^m=65 3^m - 65 = 2^m The first exponent of 3 which results in a number greater than 65 is 4 so 3^4 = 81 81-65 = 2^m 16 = 2^m 16 can be written as 2^4 m=4
You cant take any value of 'm'.consider the question is same but with a very large value(instead of 65),probably in crores,it wout take an eternity to reach that number
I was impressed by the analytic demonstration used to figure out that m = 4. Sometimes procedures can be more interesting to follow along than just knowing the result.
Well, you'll be impressed to see that the proof is not valid. If you suppose that m is an even number then x+y and x-y are integers but if it's an odd number then they are IRRATIONALS and you can't use 65=5*13. Plus the fact that even in the case where x+y and x-y are integers, 5*13 is not the only way to get 65, you must also look at 1*65...
@@italixgaming915This looks like diophantine method used here, which would work only with the integers. In case the solution works like in above case it could prove that no other integer solutions exist... But it seems 65x1 was not checked. This could work for 3^m-2^n cases also. Variables 'm' and also 'n' are often used for integers, but not necessarily always.
the proof is not valid, we need to proceed much more cautiously with analytic ways, i d say : some more conditions/discussions needed to be added to the video .... i gree with italixgaming the arithmetic way stays safer ..
65=13•5=(9+4)×(9-4)=9^2-4^2=3^4-2^4 4 is the m; My high school math teacher used to tell me,the easiest way to understand an equation is to make them look the same,that is to say,we should make the brief side more complicated other than simplifing the complicated side for the most of the time.
There is a problem here, when you write (x+y)(x-y)=5x13 you can not deduce that x+y=13 because you do not know if x+y is a natural number. If m is an odd number, then x=3^(m/2) and y=2^(m/2) are not natural numbers. So you have prove that if m is an even number, then m=4. it is very easy to check that there is only 1 solution.
@rubikaz "So you have prove that if m is an even number" - Or, you can assume that m is even (thus making the quantities x+y and x-y (positive) integers etc.) and see whether it pays off - which it does, actually. "it is very easy to check that there is only 1 solution." - Indeed. The uniqueness of the solution is a direct consequence of the properties of the exponential function at work here...
@@babetesfaye1001 the function f(x) = 3^x - 2^x where x>0 is strictly growing, therefore with x=6 , f(6) > 65 so x must be less than 6, and so on trying integers until finding x=4 or m=4
You can also factor 65 into 65 and 1. This gives values of a and b as 33 and 32 hence b=2^(m/2)=32, m= 10 but m will have different value for 3^(m/2)= 33. Using logs(or ln) m=(log 33/log 3)=6.365
@samuelmayna "You can also factor 65 into 65 and 1." - Yes, you can, but these won't be proper factors for 65, in the sense that ANY NUMBER could be "factored" as itself and one... Also, as you've seen yourself, this breaks the consistency of the original equation by forcing the unknown to take different values simultaneously - which is obviously not possible.
I got ‘m=4’ by mental arithmetic. Because 3^m > 65 and 2^m < 3^m, that’s necessary. The value '65' determines that the value range of m must be less than 5 and greater than 0. When m is a positive integer, test the m=5, 4, 3, 2, 1 and finally get m=4.
Since we don't participate in the Olympics, this short answer is good. But I'm not sure if I would use this method for an Olympics. I think logarithms are simpler than the answer And have a good axiom to condense those complicated answers.❤
exactly! if students are good enough to do olympiad problems then they'd know powers of 3, 3, 9, 27, 81, 243 at least and also 2 to even higher powrers, 2, 4 , 8, 16, 32 etc so very easy to work out in less than 20 seconds . However I'm used to doing mental arithmatic as I never had a calculater when I was in school
Similarly you can as well re-write 65 as 81-16....From there you make the bases of the two numbers to be similar with what you have on the left hand side..From there you take one of the corresponding bases and equate them together,when bases are the same powers will also be the same.
you are not mathematically solving the problem, but doing so by trial and error. These were smaller numbers so it is easy for anyone to come to that conclusion ( becausethe solution is "visible" in the numbers in front of you). In other words, how would you solve the same problem with entirely different and larger numbers involved? ...say 2059 for instance.
Ευχαριστώ πολυ γι αυτες τις λυσεις...ειμαι 65 ετων και μου αρέσει να θυμαμαι τα χρόνια μου στο σχολείο...Παντα μου αρεσαν τα μαθηματικα και τωρα τα θυμάμαι ολα και οτι δεν θυμαμαι το μαθαινω μαζι σας thank you❤
@mitahaubica6498 "I immediately saw that 65 can be decomposed as 81-16" - That's not decomposition... You can find an infinity of pairs of numbers that have their difference equal to 65. The fact that you selected one such pair that also happens to be powers of the respective bases in the original problem merely indicates you have approached solving this by trial and error... *The question would be whether you can do better than finding the solution that way...*
How can one just assume that x+y = 13, and x-y = 5, respectfully, as there's 65 and 1 as well. Furthermore, this wouldn't really work if 65 had many more factors, making it more complicated, opening up to a lot more possibilities.
Nice question @Santhosh John. There are principles and rules that govern the operations in mathematics as it is for all other sphere of life. Once you are a maths student or tutor you must get urself familiarized with these rules. They become part of you and you know what to do once you have a math challenge/problem before you. Once a mathematician sees a math problem, his head automatically runs through different means of approaching the problem for a better solution.
When taking difference of squares if smaller part is 1(assuming it is the smaller part and it is) , m=2 and positive part becomes 5, not 65 therefore the result will be 5, not 65. If it has more factors, you just have to make arithmetical inferences and simplify it. That's it.
@@naharmath but there is no other solution since both parts are exponential and even if m is a rational number the result wouldn't be integer( they are different primes). So this equation requires to be analyzed numerically first
Il professore ti ha risposto in maniera adeguata. Però io consiglio al professore di spiegare certe regole anche se ciò richiede qualche minuto in più. I fruitori di you tube non sono tutti matematici, ma persone desiderose di capire ed imparare.
The way I see it, 3 cubed is 27, less than 65, and 3 raised to 4 is 81. Therefore 65 is between 27 and 81. Upper bound, lower bound or range even. Now 81 would mean m is 4. So 2 should be raised to 4 also which gives us 16. 81 minus ➖ 16 is 65. So m is 4. The other ways are using log or binomial series which is overkill for smaller numbers. - Amy
Another option: the solution is pretty clear, it's a small number one can calculate with the mind. Then it reduces to show that their solution is unique. One can use calculus to show that the analog continuos function is monotone for x>4 and be done with it, or use induction to show that it grows on the integers for n>4.
Your approach is superb sir. I find it very fascinating and I will try it out in subsequent math challenges. Thanks for sharing this nice and wonderful procedure with OnlinemathsTV. You the boss and much respect boss...👍👍👍
I think he used a simplified equation for the demonstration. When it's not so simple and the numbers are much larger, then this method can be used as well.
Thank you for reminding me that you can't believe everything on the internet. Props to you i almost believed it untill i tested it with calculator. BRAVO YOU SHOULD WIN AN OSCAR
Parabéns. Não sei falar nada em inglês e mesmo assim consegui aprender com sua aula. Até eu estou surpreso de ter assistido sua aula até o final sem saber se iria entender o seu modo de esnsinar. A matemática pode ser universal, mas o jeito de ensinar é fundamental.
@Airtonreis, we want to sincerely say you are such a wonderful person and thanks a million for watching our contents despite the language barrier. We all @OnlineMathstv deeply cherish and love you from the depth of our hearts sir. ❤️❤️❤️💖💖💖💕💕💕🙋🙋🙋
All I did was think whats the first interger power of 3 that goes past 65, 3 is 27 so the answer is 4. Then just test. and it gave right answer. simple.
@@onbored9627 Where are you from ? Cause how did you know I was still alive after 4 months ? 🙂Here is a secret about me : I 've never studied Mathematics in English !
@@Quasar900 Ah, I apologize I only speak English. I'm from the USA. I wasn't trying to take away from your explanation I should've been more clear on that, I just thought yours was so good I didn't even need to say. Figuring out it's strictly growing is clever as hell. I didn't even think of that. I meant simple, as in, my answer was simply a guess really and it worked.
@@onbored9627 Oh please Sir , no need to Apologise , The fact that I've never studied math in English doesn't mean I don't know English 🙂 It's just I'm not that familar with English terms in math ! But Thank God I do speak and read : French, English, Arabic, Spanish + some Japanese ! I did study math in French (after high school) and Arabic (until high school) , but that waaaas 21 years ago ! What class are you in ? I hope you're safe from those ongoing blizzard storms ! Greetings From Morocco and Free Palestine 🙂
@@onbored9627 I think you do know these techniques involving the continuity of a function, the intermediate values etc.. to solve equations ! For example : solving in IR set : Arctan(x+1)+Arctan(x-1)=Pi/4
I like how you used indices to get past that first section. I've never really been good at spotting where to use substitution, the section when you brought in x and y
If you guess the solution, m=4 and present the equality as 3^m - 81 = 2^m - 16 , then it would be easy to prove that both functions (on the right and on the left) increase and therefore their graphs have only one intersection.
You need more than that. Two increasing functions can intertwine and cross each other all the time. But if they have one intersection, and after that one of them grows faster than the other all the time, then it follows that they won't intersect again. It's like two cars racing. They both increase their distance from the start all the time, but they could swap places many times during the race--unless one of them is always faster than the other after the point in which they were even with each other.
@@reginaldocalvo4361Each of the two sides can be a function e.g. y = 3^x - 81. The solution of the equation 'side 1' = 'side 2' is a number x (or m) where the two functions give the same y for the same x. You first find, by guessing, one x (or m) for which the two functions have the same y, which would mean that the particular m is one solution to the equation of function 1 = function 2 for some x. You then show that each of the functions only grows, and that the difference between the two y-s for each x (which is a function of x as well) also only grows. Therefore there can be only one value of x for which the difference is 0, so only one m (the one we already guessed) is the solution to the equation where one of the functions has the same y as the other for a given x. If I were participating in this Olympiad and solving this problem, I would have guessed 4, and then I would have argued that the first side 3^m... grows much faster than the other side 2^m for each subsequent m. I wouldn't be using derivatives, as I did not know any calculus in high school. But if they ask about integer solutions, I would talk about growth of y with respect to changes in m by +1. And if they did not limit it to whole numbers, I would probably still try to talk about slope of the graphs of the functions and hope to make an acceptable argument, as I don't see how you can analyze these functions without using calculus.
We are glad you love what is happening here. We promise to give more educative contents in the area of mathematics with the help of God. Love you.....💖💖💕💕
I have been learning Genetic Algorithms in Python; they are good for problems like this. The value I get for 'm' is 3.97, 3.99, 4.00, etc.; different each time as there is a random element for the convergence. A little intuition is also required; If you plug 4.0 into the equation you get correct 65. Program run time ~5 seconds.
Hello. I program in PureBasic. I made this task by binary search method, 30 iterations, result 3.9999999991, precision 0.0000000047. Time is almost instantaneous.
Very well solution but the rules are very lengthy or Expensive ☺️ So if we assume the value of "m" here from 1-4 We are easily getting the value of m Such as Let , m=1,2,3,4,... ♾️ and now, 3^1-2^1=1≠65 again 3^2-2^2=5≠65 And now if we let, m=4 then 3^4-2^4=65=65 So we can easily get m=4😊
I did it in a different way. Write 3 as 2+1 and perform binomial expansion, so the 2^m cancels and subtract one from both sides. We have 64 on one side and some series on other side. Notice that the series 2 + 2^2 +.....+ 2^m will definitely be smaller than series on left which is equal to 64. So make this G.P. sum less than 64, you will get that m should be less than 5, and once you have known this, you have proved that you just need to find a solutions less than 5 and those will be the only solutions. Only m=4 works out.
If you do the analysis, and supposing m is an integer, you can conclude m is even. That's because 3^m-2^m must be congruent to 0 mod (5). If m is even you get 1 mod 5 or 4 mod 5. But m being even, you have 0 mod 5 always. Anyways, it's not proven in the video, maybe it would be amazing to have hows and whys in the video
It can decide differently. The function (1) f(x)=3^x-2^x -is increasing . { for x>0 (2) 3^x>2^x ; x10 . (3) f(x2)-f(x1)= ……..=3^x1*[3^(x2-x1)-1]-2^x1*[2^(x2-x1)-1 ]> 2^x1*[2^(x2-x1) -1 ]-[ “--“ ]=0 ; (3) f(x2)-f(x1)>0 !!!!! So , it takes all its meanings once a time. f(4)=3^4-2^4=65 . So , x=4 - is the only root of equation ! Respectfully , Lidiy
The elders in our mist are highly learned. Love your detailed explanation sir. Thanks for finding our time to watch our content and commenting even at this age of yours sir. Much respect and we All @onlinemathstv love you dearly...💖💖💕💕
Pay attention that m=constant not variable, so M must be grater then 1 otherwise solution will not be true because ln1=0 or 3-2=1 which is not equal 65. So M>1 and could be anything. So if it is not an integer number? So if it is not equal 65 but 63.5 - ?
I think it's very easy. We let m = 4, it's correct We will prove "with m > 4 , 3^m -2^m > 65" So, only m = 4 will be equation. m>4 m = 4+a (a>0) 3^m - 2^m > 3^4 - 2^4 Because (3^a-1).3^4 always > (2^a-1).2^4 Thanks from Việt Nam 🎉
Rewrite the equation as 3^x = 65 + 2^x. We can safely divide by 3^x and then we have that 1 = 65 * (1/3)^x + (2/3)^x. Because the function on the R.H.S. is strictly decreasing, being the sum of 2 other strictly decreasing functions, it means that f(x) = 1 has one solution at max. We notice that x=4 checks, so that is our only solution.
Sir, we can also do this by hit and trial method, assuming different values for m=1,2,3,4.... But your way was also very nice 👍👍. Thank you for this video sir🙏
Yes, but that will only work if the working process is not in the examination but this is needed when the examiner wants you to show your procedure step by step...👍👍👍
@@gregfarnham5651 The solution in the video also made use of the assumption that m is an integer. Otherwise the factorization of 65 would not be unique. Actually, the hit and trial method should be entirely ok.
@@dilphekyou guys got a different Olympiad or something? Here Olympiads (for the kids, totally different type from the subjective ones) are mcqs and the subjective ones are like 3 questions in 3 hrs and if you're able to do even 1 you're qualified (you can imagine the toughness so it's really not for the kids)
Yes it has multiple answer but for the sake of this tutorial we restricted ourselves to this solution sir. Thanks for this observation. Much love....💕💕👍👍
Clever solution! One thing I did not get is how you knew that m was a positive integer - or was this just an assumption that happened to work? For example, if the problem had been 4^m - 3^m = 65, m would be approximately 3.36.
Слева возрастающая функция при m>0 ( можно найти производную и убедиться ), справа постоянная функция, значит у них может существовать только одна точка пересечения, методом оценки m=4
There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in: 3^m = 65. m is greater than 3, because 3^3 = 27. m can be 4 because 3^4 = 81. Let's test m = 4: 3^4 - 2^4 = 65 81 - 16 = 65 65 = 65
Wow!!! This is fantastic. I love this approach sir. We have gained some values from this procedure sir. Thanks for dropping this sir. Respect.....👍👍👍 Much love....💕💕💖💖❤️❤️
There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in: 3^m = 65 m = log(65)/log(3) = 3,8 m is greater than 3,8 Let's test m = 4 3^4 - 2^4 = 65 81 - 16 = 65 65 = 65
Base on which level you’re at. This question and video are made for yr 10-11? So he gave the solution at that grade. Above yr 12 can use other tools as log/ ln skilfully to solve it.
Before watching: Alright, so, we are looking at exponential functions. 3^m - 2^m = 65. First, we can rule out m=1 and anything below; the difference between those would be smaller than between 3 and 2, and thus much lower than 65. Next, notice that we are dealing with an integer on the right. This heavily implies (but *does not necessarily guarantee* ) that 3^m and 2^m are both integers as well. If both are integers, then m must also be an integer. So we should start with integers. (If we were dealing with a mixed number instead, this would be much more complex; as it stands, we can just plug in integers and see which one works). Let's start with the first x that gives us 3^m >65, namely 4. 3^4 = 9^2 = 81, and 2^4 = 4^2 = 16 So for m= 4, we have 81-16 = 65. Fortunately for us, this checks out, and thus we have our answer of *m = 4*
@@RoderickEtheria Yes, you are correct, and I see I made a typo in the "so for m=4 we have..." section by putting a negative in front of the 4. Fixed now. I don't imagine it was a huge problem since in the section right above it,and at the very end of that section, I used 4, but it still may have confused some people.
Сначала поделить обе части на 2^m. Тогда слева будет возрастающая функция, справа убывающая. Тогда уравнение имеет не более одного корня. Подобрать корень не сложно. 9 класс, ничего сложного. За проведенное решение минус: нигде не доказано отсутствие других решений, переход к системе ничего не обосновывает.
This is so overly complicated.. Just check low values of m and find that m = 4 works.. Then use a simple analysis tool to show uniqueness of the solution e.g. by showing that function f(m) = 3^m - 2^m is strictly increasing for m>=1. Additionally your solution contains errors and missing steps: 1. If you do the trick with 3^(m/2)^2 - 2^(m/2)^2 = 65 then you presuppose that m is even, because if m is odd then 3^(m/2) is not natural anymore, so you cannot use the natural divisors of 65 in the following steps anymore. 2. Even if that approach worked (because you somehow proved that m must be even): After you rewrite the equation as x^2 - y^2=65, then you have to consider *two* pairs of solutions 65 = 65 * 1 and 65 = 5 * 13. Long story short, lots of mistakes in your video unfortunately.
Thanks for this keen observation and I really appreciate this comment sir. Noted. I will do more detailed work on subsequent videos. You the best and much love for this detailed comment....💕💕💕
Yes this video is shit full of errors. The fact that it has so many errors and is hugely overcomplicated at the same time makes it complete and utter shit
3ᵐ-2ᵐ=65 y=3ˣ-2ˣ y=65 если построить оба эти графика, то будет видно, что уравнение имеет единственное решение, поэтому можно попробовать подобрать корень подбором: х=1: 3¹-2¹=1; 1≠65 х=2; 3²-2²=5; 5≠65 х=3; 3³-2³=19; 19≠65 х=4; 3⁴-2⁴=65; 65=65 х=4 корень m=4
in maths olympiad time is very critical...what is important is the answer...so the best way is to solve maths olympiad is to have very basic maths then the rest is analysing to get the pattern...so what i did is just look at indecies of 3 that have the last number such that if we subract an indecies of 2 and get 65
if m0 and the derivative of 3^m-2^m is 3^m*log(3) - 2^m*log(2) which in turn larger than 3^m*log(2)-2^m*log(2) which is clearly positive as both log(2) > 0 and 3^m>2^m hold. Therefore the solution is unique, since 3^m-2^m-65 is increasing. A simple guess shows m=4 is a solution.
The presentation and demonstrated solution is very good. rubikaz comment is nevertheless valuable. If the approach solves, then the uniqueness provides all solutions. Using an exclusive deduction from the start is valid, if a working hypothesis. Thanks for this video.
I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚
1. (3^m-2^m) isincreasing function of m. 2.M is integer,because 65 is integer. 3.f(0)=0,f(5)=243-32=211. 4.we only need to test by 1,2,3,4 5.The answer is 4
You worked hard, I found the result by giving the value of m to 4 in 10 seconds, but it is important how the solution is, but I still think it is too long. Greetings from Turkey.
So helpful . And this video gave me idea how to solve elzebric brobepems ❤ . Take more video related to solve many brobepems in polynomial . Logorithimc function and trigonometry Nice and very helpful video
Thanks a million for watching and dropping this wonderfully encouraging comment sir. The true is that mathematics is all about finding a solutions to problems/challenges. We really appreciate the fact that you watch and dropped this comment to clarify some debts in the minds of so many viewers and subscribers here. Much love from all of us @OnlinemathsTV....💕💕💕💖💖💖❤️❤️❤️
Basit düşünecek olursak 3^m ve 2^m degerleri fonksiyon oldugundan fonsiyonlarda m icin koordinat düzlemi mantigiyla hareket edersek bu tip sorularda yapilacak en kestirme yol m degerleri yerine m=0,1,2,3,4,5,... Ve negatif taraflardan -1,-2,-3,-4,-5,... Degerlerinden verilince direkt cevap cikar. M= 4 direkt cevap cikar. Birkac denemeden sonra direkt cevap çıkması olasıdır. (Fonksiyonda noktayi yazip sonucu bulmak gibi). Bircok soruda direkt cevap cikar. (Bircok sorunun cevabı da iki kare farkı, toplamın karesi, farkın karesinden direkt cevap çıkar.). Bircok olimpiyat sorusu da bu metotla çözülebilir. Tamami degil ama sorularin çözümlerinde saglam, kısa süreli çözersiniz. Öğretmenin emegine saglik, cok güzel anlatmış bu arada
Before watching this video i want to share my method - By using binomial expansion 3^m - 2^m =65 =>(1+2)^m - (1+1)^m =65 =>(1+2m + (2m(2m+1)/2!) +........) - (1+m+(m(m+1)/2!)+.......)= 65 (by binomial expansion) Bro i got stuck 😮💨
I was able to solve the problem in less than 1 minute... I just thought about the Multiple of 3 more than 65 that was 81 and then i subtracted 65 from it and the resulting answer was 16,now we know that 3^4 is 81 and 2^4 is 16,so easily the value of m came out to be 4
The comment section is polluted by critics but i learned some tricks from this video.. EXCELLENT
Wow!!! Thanks and we are glad you gained some values from this video tutorial sir.
it isnt critism.. in every maths channel you will see people doing that.. they do it for alternate solutions and writing down their own way
@@thefireyphoenix this video wasn't made for them
Polluted by critics??
You're definitely not a student of maths
I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚
I worked it out a bit different. My solution was simply determine what of 3 exponent would get me a number greater than 65 that would be an odd number (3^4). I then subtracted that 65 from that number (81) and I got 16 which is 2^4.
In other words you can rewrite the equation in this instance as
3^m - 2^m=65
3^m - 65 = 2^m
The first exponent of 3 which results in a number greater than 65 is 4
so 3^4 = 81
81-65 = 2^m
16 = 2^m
16 can be written as 2^4
m=4
That's exactly what I did.
m can be a negative number?
You cant take any value of 'm'.consider the question is same but with a very large value(instead of 65),probably in crores,it wout take an eternity to reach that number
I did the same😂
You're assuming that m is an integer?
I was impressed by the analytic demonstration used to figure out that m = 4. Sometimes procedures can be more interesting to follow along than just knowing the result.
Well, you'll be impressed to see that the proof is not valid. If you suppose that m is an even number then x+y and x-y are integers but if it's an odd number then they are IRRATIONALS and you can't use 65=5*13.
Plus the fact that even in the case where x+y and x-y are integers, 5*13 is not the only way to get 65, you must also look at 1*65...
@@italixgaming915This looks like diophantine method used here, which would work only with the integers.
In case the solution works like in above case it could prove that no other integer solutions exist... But it seems 65x1 was not checked.
This could work for 3^m-2^n cases also. Variables 'm' and also 'n' are often used for integers, but not necessarily always.
You are the top!😊 You are a great teacher!!!!
the proof is not valid, we need to proceed much more cautiously with analytic ways, i d say : some more conditions/discussions needed to be added to the video .... i gree with italixgaming
the arithmetic way stays safer ..
Support
65=13•5=(9+4)×(9-4)=9^2-4^2=3^4-2^4
4 is the m;
My high school math teacher used to tell me,the easiest way to understand an equation is to make them look the same,that is to say,we should make the brief side more complicated other than simplifing the complicated side for the most of the time.
Wow wish I knew this when I was at school
There is a problem here, when you write (x+y)(x-y)=5x13 you can not deduce that x+y=13 because you do not know if x+y is a natural number. If m is an odd number, then x=3^(m/2) and y=2^(m/2) are not natural numbers. So you have prove that if m is an even number, then m=4. it is very easy to check that there is only 1 solution.
Noted.
@rubikaz "So you have prove that if m is an even number" - Or, you can assume that m is even (thus making the quantities x+y and x-y (positive) integers etc.) and see whether it pays off - which it does, actually.
"it is very easy to check that there is only 1 solution." - Indeed. The uniqueness of the solution is a direct consequence of the properties of the exponential function at work here...
❤❤❤
May I use any number Which have same numerator band denominator ? like 3/3 , 4/4 , 5/5....
@@babetesfaye1001да, потому что оно равно 1
@@babetesfaye1001 the function f(x) = 3^x - 2^x where x>0 is strictly growing, therefore with x=6 , f(6) > 65 so x must be less than 6, and so on trying integers until finding x=4 or m=4
You can also factor 65 into 65 and 1. This gives values of a and b as 33 and 32 hence b=2^(m/2)=32, m= 10 but m will have different value for 3^(m/2)= 33. Using logs(or ln) m=(log 33/log 3)=6.365
This makes sense and it is mathematically correct
@samuelmayna "You can also factor 65 into 65 and 1." - Yes, you can, but these won't be proper factors for 65, in the sense that ANY NUMBER could be "factored" as itself and one... Also, as you've seen yourself, this breaks the consistency of the original equation by forcing the unknown to take different values simultaneously - which is obviously not possible.
m is an integer
@@nicadi2005 but it is mathematically logical.Mathematics is about thinking all cases.
@@AtefFarrouki but my approach is sound which shows that 65 and 1 won't work but it can give different answers for some equations.
I got ‘m=4’ by mental arithmetic. Because 3^m > 65 and 2^m < 3^m, that’s necessary.
The value '65' determines that the value range of m must be less than 5 and greater than 0. When m is a positive integer, test the m=5, 4, 3, 2, 1 and finally get m=4.
I too
Since we don't participate in the Olympics, this short answer is good. But I'm not sure if I would use this method for an Olympics. I think logarithms are simpler than the answer And have a good axiom to condense those complicated answers.❤
I dont think so. 2^m 65 why m
Me too! 😊😊
exactly! if students are good enough to do olympiad problems then they'd know powers of 3, 3, 9, 27, 81, 243 at least and also 2 to even higher powrers, 2, 4 , 8, 16, 32 etc so very easy to work out in less than 20 seconds . However I'm used to doing mental arithmatic as I never had a calculater when I was in school
Similarly you can as well re-write 65 as 81-16....From there you make the bases of the two numbers to be similar with what you have on the left hand side..From there you take one of the corresponding bases and equate them together,when bases are the same powers will also be the same.
This is what I actually expected from him
If you do not notice that 16 = 2**4, you are not a computer geek.
I did it,in my braine.
th-cam.com/video/sIlRHg730CA/w-d-xo.htmlsi=ujk2KD_xjoMGqiP5
you are not mathematically solving the problem, but doing so by trial and error. These were smaller numbers so it is easy for anyone to come to that conclusion ( becausethe solution is "visible" in the numbers in front of you). In other words, how would you solve the same problem with entirely different and larger numbers involved? ...say 2059 for instance.
Ευχαριστώ πολυ γι αυτες τις λυσεις...ειμαι 65 ετων και μου αρέσει να θυμαμαι τα χρόνια μου στο σχολείο...Παντα μου αρεσαν τα μαθηματικα και τωρα τα θυμάμαι ολα και οτι δεν θυμαμαι το μαθαινω μαζι σας thank you❤
Damn you fasho deserve my sub
I immediately saw that 65 can be decomposed as 81-16, and conveniently 81 is 3^4 and 16 is 2^4, so matching coefficients suggests m is 4.
You are a genius
@mitahaubica6498 "I immediately saw that 65 can be decomposed as 81-16" - That's not decomposition... You can find an infinity of pairs of numbers that have their difference equal to 65.
The fact that you selected one such pair that also happens to be powers of the respective bases in the original problem merely indicates you have approached solving this by trial and error...
*The question would be whether you can do better than finding the solution that way...*
that is not the right method..may work here but not al the time
You just got lucky. It was pure luck that you used the right numbers to subtract, and that m is an integer in this case.
Always try some values of m to see the behaviour.
m=0 or 1 or 3 or 4... I've found the solution!
Obviously it is not an Olympiad problem.
How can one just assume that x+y = 13, and x-y = 5, respectfully, as there's 65 and 1 as well. Furthermore, this wouldn't really work if 65 had many more factors, making it more complicated, opening up to a lot more possibilities.
Nice question @Santhosh John. There are principles and rules that govern the operations in mathematics as it is for all other sphere of life.
Once you are a maths student or tutor you must get urself familiarized with these rules.
They become part of you and you know what to do once you have a math challenge/problem before you.
Once a mathematician sees a math problem, his head automatically runs through different means of approaching the problem for a better solution.
When taking difference of squares if smaller part is 1(assuming it is the smaller part and it is) , m=2 and positive part becomes 5, not 65 therefore the result will be 5, not 65.
If it has more factors, you just have to make arithmetical inferences and simplify it. That's it.
3^(m/2) is not necesserly an integer!
@@naharmath but there is no other solution since both parts are exponential and even if m is a rational number the result wouldn't be integer( they are different primes). So this equation requires to be analyzed numerically first
Il professore ti ha risposto in maniera adeguata. Però io consiglio al professore di spiegare certe regole anche se ciò richiede qualche minuto in più. I fruitori di you tube non sono tutti matematici, ma persone desiderose di capire ed imparare.
Maybe it's a bad argument but I would say for a^m - b^m = X, and a, b, m belong to N, a^m> X > a^m-1.
Here 3^m>65>3^m-1
81>65>27 => m=4.
The way I see it, 3 cubed is 27, less than 65, and 3 raised to 4 is 81. Therefore 65 is between 27 and 81. Upper bound, lower bound or range even. Now 81 would mean m is 4. So 2 should be raised to 4 also which gives us 16. 81 minus ➖ 16 is 65. So m is 4. The other ways are using log or binomial series which is overkill for smaller numbers. - Amy
You tried here tutor Jakes. I have learnt something here. Just keeping on running this channel. More grace, love from Port Harcourt ❤.
It is our pleasure to serve you sir. Thanks for watching
Thankyou you,i understand😊
Another option: the solution is pretty clear, it's a small number one can calculate with the mind. Then it reduces to show that their solution is unique. One can use calculus to show that the analog continuos function is monotone for x>4 and be done with it, or use induction to show that it grows on the integers for n>4.
Your approach is superb sir. I find it very fascinating and I will try it out in subsequent math challenges.
Thanks for sharing this nice and wonderful procedure with OnlinemathsTV.
You the boss and much respect boss...👍👍👍
I think he used a simplified equation for the demonstration. When it's not so simple and the numbers are much larger, then this method can be used as well.
A matemática é uma língua universal como a música. Parabéns, ótima técnica ❤
Thank you for reminding me that you can't believe everything on the internet. Props to you i almost believed it untill i tested it with calculator. BRAVO YOU SHOULD WIN AN OSCAR
Parabéns. Não sei falar nada em inglês e mesmo assim consegui aprender com sua aula. Até eu estou surpreso de ter assistido sua aula até o final sem saber se iria entender o seu modo de esnsinar. A matemática pode ser universal, mas o jeito de ensinar é fundamental.
@Airtonreis, we want to sincerely say you are such a wonderful person and thanks a million for watching our contents despite the language barrier.
We all @OnlineMathstv deeply cherish and love you from the depth of our hearts sir. ❤️❤️❤️💖💖💖💕💕💕🙋🙋🙋
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Brasil tá em todo lugar não tem jeitoooooo
the function f(x) = 3^x - 2^x where x>0 is strictly growing, there for if x=6 f(x) > 65 so x must be
All I did was think whats the first interger power of 3 that goes past 65, 3 is 27 so the answer is 4. Then just test. and it gave right answer. simple.
@@onbored9627 Where are you from ? Cause how did you know I was still alive after 4 months ? 🙂Here is a secret about me : I 've never studied Mathematics in English !
@@Quasar900 Ah, I apologize I only speak English. I'm from the USA. I wasn't trying to take away from your explanation I should've been more clear on that, I just thought yours was so good I didn't even need to say. Figuring out it's strictly growing is clever as hell. I didn't even think of that. I meant simple, as in, my answer was simply a guess really and it worked.
@@onbored9627 Oh please Sir , no need to Apologise , The fact that I've never studied math in English doesn't mean I don't know English 🙂 It's just I'm not that familar with English terms in math !
But Thank God I do speak and read : French, English, Arabic, Spanish + some Japanese !
I did study math in French (after high school) and Arabic (until high school) , but that waaaas 21 years ago !
What class are you in ? I hope you're safe from those ongoing blizzard storms !
Greetings From Morocco and Free Palestine 🙂
@@onbored9627 I think you do know these techniques involving the continuity of a function, the intermediate values etc.. to solve equations !
For example :
solving in IR set :
Arctan(x+1)+Arctan(x-1)=Pi/4
I like how you used indices to get past that first section. I've never really been good at spotting where to use substitution, the section when you brought in x and y
Hahahaha....thanks a bunch my good friend and thanks for watching our contents consistently.
We all here love deeply....💖💖💕💕😍😍
If you guess the solution, m=4 and present the equality as 3^m - 81 = 2^m - 16 , then it would be easy to prove that both functions (on the right and on the left) increase and therefore their graphs have only one intersection.
You need more than that. Two increasing functions can intertwine and cross each other all the time. But if they have one intersection, and after that one of them grows faster than the other all the time, then it follows that they won't intersect again. It's like two cars racing. They both increase their distance from the start all the time, but they could swap places many times during the race--unless one of them is always faster than the other after the point in which they were even with each other.
@@reginaldocalvo4361Each of the two sides can be a function e.g. y = 3^x - 81. The solution of the equation 'side 1' = 'side 2' is a number x (or m) where the two functions give the same y for the same x. You first find, by guessing, one x (or m) for which the two functions have the same y, which would mean that the particular m is one solution to the equation of function 1 = function 2 for some x. You then show that each of the functions only grows, and that the difference between the two y-s for each x (which is a function of x as well) also only grows. Therefore there can be only one value of x for which the difference is 0, so only one m (the one we already guessed) is the solution to the equation where one of the functions has the same y as the other for a given x.
If I were participating in this Olympiad and solving this problem, I would have guessed 4, and then I would have argued that the first side 3^m... grows much faster than the other side 2^m for each subsequent m. I wouldn't be using derivatives, as I did not know any calculus in high school. But if they ask about integer solutions, I would talk about growth of y with respect to changes in m by +1. And if they did not limit it to whole numbers, I would probably still try to talk about slope of the graphs of the functions and hope to make an acceptable argument, as I don't see how you can analyze these functions without using calculus.
@@reginaldocalvo4361
3^m-81= f(m) this can be considered as a function depending on m
2^m-16= g(m) also could be considered as function of m.
@@ca1498 you are right, but
When you see this function it is easy to notice that the growing path or direction is already know but partially
Consequence is false.Contra-example x and x^3. Both increase but have 2 intersections
Love your video! I got lost half way and I did not know that you could just square the exponents and make them equal
We are glad you love what is happening here.
We promise to give more educative contents in the area of mathematics with the help of God.
Love you.....💖💖💕💕
I like this too. I have done it with roots before. Roots are just indices but I wouldn't have thought of doing it.
Очень грамотное изложение, и очень удобно следить на доске. Однозначно плюс!🎉
I have been learning Genetic Algorithms in Python; they are good for problems like this.
The value I get for 'm' is 3.97, 3.99, 4.00, etc.; different each time as there is a random element for the convergence.
A little intuition is also required;
If you plug 4.0 into the equation you get correct 65. Program run time ~5 seconds.
Wow!!! Nice sir.
If you're going the numerical route, bisection search is much faster.
Blunt enumeration of roots is not always the best solution)
Hello. I program in PureBasic. I made this task by binary search method, 30 iterations,
result 3.9999999991, precision 0.0000000047. Time is almost instantaneous.
You need to review the Python solutions cos only 4 is an exsct solution. 3.97 is far from it, 3.99 is just an approximate
Very well solution but the rules are very lengthy or Expensive ☺️
So if we assume the value of "m" here from 1-4
We are easily getting the value of m
Such as
Let , m=1,2,3,4,... ♾️
and now,
3^1-2^1=1≠65
again
3^2-2^2=5≠65
And now if we let, m=4 then
3^4-2^4=65=65
So we can easily get m=4😊
Bravo👍👍👍
i mean hit and trial is always the last resort
Great job, You are an excellent instructor. From an admirer Ethiopian in America
I did it in a different way.
Write 3 as 2+1 and perform binomial expansion, so the 2^m cancels and subtract one from both sides. We have 64 on one side and some series on other side.
Notice that the series 2 + 2^2 +.....+ 2^m will definitely be smaller than series on left which is equal to 64.
So make this G.P. sum less than 64, you will get that m should be less than 5, and once you have known this, you have proved that you just need to find a solutions less than 5 and those will be the only solutions.
Only m=4 works out.
a good method to find the range of m and it makes sensible checking easier with limited number of m
Yeah
In a slightly fancier approach we can make a recursion
2(3^m - 2^m) + 3^m = (2+1)3^m - 2^(m+1)
= 3^(m+1) - 2^(m+1)
Put a(m) = 3^m - 2^m
a1 = 3^1 - 2^1 = 1
a2 = 2 a1 + 3^1 = 5
a3 = 2 a2 + 3^2 = 19
a4 = 2 a3 + 3^3 = 65, done
Wow!!! This approach is impressive but a bit obscure sir.
can u explain what u mean by obscure pls😂
can you help me sir?
if ab+bc+ca=1
prove:
sqrt(a + (1/a)) + sqrt(b + (1/b)) + sqrt(c + (1/c)) >= 2( sqrt(a) +sqrt(b) + sqrt(c) )
I did it in my mind within 4 seconds by putting natural numbers.
It doesn't work if m is odd. In this case (x+y) and (x-y) aren't integer and can't be assumed as 5x13.
If you do the analysis, and supposing m is an integer, you can conclude m is even.
That's because 3^m-2^m must be congruent to 0 mod (5). If m is even you get 1 mod 5 or 4 mod 5. But m being even, you have 0 mod 5 always.
Anyways, it's not proven in the video, maybe it would be amazing to have hows and whys in the video
@@danielrivera2278 of cause. I mean that "trick" in the video is not universal.
@@danielrivera2278 and who even said that m must be an integer ?
@@Change_Verification exactly. I also think like that, that's because assuming integer was the first thing I said
Yes, he didn't do the preliminary work. But this is an important step to make tricks like this one work.
It can decide differently. The function (1) f(x)=3^x-2^x -is increasing . { for x>0 (2) 3^x>2^x ; x10 . (3) f(x2)-f(x1)= ……..=3^x1*[3^(x2-x1)-1]-2^x1*[2^(x2-x1)-1 ]> 2^x1*[2^(x2-x1) -1 ]-[ “--“ ]=0 ; (3) f(x2)-f(x1)>0 !!!!! So , it takes all its meanings once a time. f(4)=3^4-2^4=65 . So , x=4 - is the only root of equation ! Respectfully , Lidiy
The elders in our mist are highly learned.
Love your detailed explanation sir.
Thanks for finding our time to watch our content and commenting even at this age of yours sir.
Much respect and we All @onlinemathstv love you dearly...💖💖💕💕
There is no X! 😂but however, good job!
Pay attention that m=constant not variable, so M must be grater then 1 otherwise solution will not be true because ln1=0 or 3-2=1 which is not equal 65.
So M>1 and could be anything. So if it is not an integer number? So if it is not equal 65 but 63.5 - ?
You took the words right out of my mouth. :)
Здравствуйте,Лидий! Не ожидал Вас здесь увидеть)
I think it's very easy.
We let m = 4, it's correct
We will prove "with m > 4 , 3^m -2^m > 65"
So, only m = 4 will be equation.
m>4 m = 4+a (a>0)
3^m - 2^m > 3^4 - 2^4
Because (3^a-1).3^4 always > (2^a-1).2^4
Thanks from Việt Nam 🎉
Rewrite the equation as 3^x = 65 + 2^x. We can safely divide by 3^x and then we have that 1 = 65 * (1/3)^x + (2/3)^x. Because the function on the R.H.S. is strictly decreasing, being the sum of 2 other strictly decreasing functions, it means that f(x) = 1 has one solution at max. We notice that x=4 checks, so that is our only solution.
Bravo!!!
You the best sir.
Maximum respect sir 🙏🙏🙏
profe en olimpiadas de las matematicas se aprende mucho. y con usted bastante. felicitaciones
Thanks a million sir, we appreciate this comment my good friend.
Thank you so much for sharing this wonderful learning for us Sir! God bless...❤❤❤
Sir, we can also do this by hit and trial method, assuming different values for m=1,2,3,4....
But your way was also very nice 👍👍.
Thank you for this video sir🙏
Yes, but that will only work if the working process is not in the examination but this is needed when the examiner wants you to show your procedure step by step...👍👍👍
Yes, trial and error could work if we assume m is a positive integer. I don't believe that was a given, however.
@@gregfarnham5651 The solution in the video also made use of the assumption that m is an integer. Otherwise the factorization of 65 would not be unique. Actually, the hit and trial method should be entirely ok.
@@ralfimuller8948 Agree. Thank you.
Also, by trial and error you can't prove that's the obly answer
Остроумно и красиво! Как и вся математика.
Thanks for this wonderful comment sir.
Love you....💕💕
Очень, очень. Надо же как можно. Удивительно!
3^m>65 use log of both sides to solve the inequality we get m>3.8 so start to check m=4 in the original equation the equation is satisfied and m = 4
It took a fraction of a minute to recognize m = 4, but as a mathematician, I did find this interesting. (Dr. Mike Ecker)
I think because it was easy question
Solved 3^m-2^m=65 by just thinking about the first whole number power of 3 above 65.
It is not about finding it. Olympiad is a school competition teaching kids to solve these problems mathematically
@@dilphekyou guys got a different Olympiad or something? Here Olympiads (for the kids, totally different type from the subjective ones) are mcqs and the subjective ones are like 3 questions in 3 hrs and if you're able to do even 1 you're qualified (you can imagine the toughness so it's really not for the kids)
I really think just like that...
I'm impressed with this explicit methodology 😊(12:16am)
If a.b=65, a and b can have infinite values. So, the tutor has just one answer where multiple answers are possible.
Yes it has multiple answer but for the sake of this tutorial we restricted ourselves to this solution sir.
Thanks for this observation.
Much love....💕💕👍👍
I think you add that, where m is an integer. It makes it complete. The only integer factors 65 has are 5 and 13. Nice solution 🎉
Sorry but this video is the most useless shit I have seen. You solved it by guessing and overcomplicated massively
@@mustaphaolunrebi8100 No, 65 has 4 factors: 1, 5, 13, 65.
The equation with 2 other factors should be explored for completeness
exercicio maravilhoso🥰🥰❤❤❤❤❤❤
Thanks a millions sir, we love you ❤️❤️💖💖💕💕😍😍
Thank you sir ,for this lesson.
Clever solution! One thing I did not get is how you knew that m was a positive integer - or was this just an assumption that happened to work? For example, if the problem had been 4^m - 3^m = 65, m would be approximately 3.36.
m can never be a negative otherwise we wont have 65but fraction
Слева возрастающая функция при m>0 ( можно найти производную и убедиться ), справа постоянная функция, значит у них может существовать только одна точка пересечения, методом оценки m=4
Bravo 👍👍👍
You the best sir.
Bravo 👍👍👍
You the best sir.
Чисто случайно подставил вместо m 4, и всё сошлось! Везёт мне
Thanks teacher. Today I've learnt something very new in maths. Am really surprised....
There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in:
3^m = 65.
m is greater than 3, because 3^3 = 27.
m can be 4 because 3^4 = 81.
Let's test m = 4:
3^4 - 2^4 = 65
81 - 16 = 65
65 = 65
Wow!!!
This is fantastic. I love this approach sir. We have gained some values from this procedure sir. Thanks for dropping this sir.
Respect.....👍👍👍
Much love....💕💕💖💖❤️❤️
There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in:
3^m = 65
m = log(65)/log(3) = 3,8
m is greater than 3,8
Let's test m = 4
3^4 - 2^4 = 65
81 - 16 = 65
65 = 65
Base on which level you’re at. This question and video are made for yr 10-11? So he gave the solution at that grade. Above yr 12 can use other tools as log/ ln skilfully to solve it.
Thanks for the info.
^you found one solution doesn t mean you found all solutions
Before watching:
Alright, so, we are looking at exponential functions. 3^m - 2^m = 65.
First, we can rule out m=1 and anything below; the difference between those would be smaller than between 3 and 2, and thus much lower than 65.
Next, notice that we are dealing with an integer on the right. This heavily implies (but *does not necessarily guarantee* ) that 3^m and 2^m are both integers as well. If both are integers, then m must also be an integer. So we should start with integers. (If we were dealing with a mixed number instead, this would be much more complex; as it stands, we can just plug in integers and see which one works).
Let's start with the first x that gives us 3^m >65, namely 4. 3^4 = 9^2 = 81, and 2^4 = 4^2 = 16
So for m= 4, we have 81-16 = 65. Fortunately for us, this checks out, and thus we have our answer of *m = 4*
Thanks for this thorough explanation. You the best. 👍👍
Negative m gets fractions.
@@RoderickEtheria Yes, you are correct, and I see I made a typo in the "so for m=4 we have..." section by putting a negative in front of the 4. Fixed now.
I don't imagine it was a huge problem since in the section right above it,and at the very end of that section, I used 4, but it still may have confused some people.
3^m - 2^m = 65
> 3^m - 2^m = 81 - 16
> 3^m - 2^m = 3^4 - 2^4
comparing both sides, it can be concluded
m = 4
First you must show that m is even positive integer
Noted sir.
Сначала поделить обе части на 2^m. Тогда слева будет возрастающая функция, справа убывающая. Тогда уравнение имеет не более одного корня. Подобрать корень не сложно. 9 класс, ничего сложного. За проведенное решение минус: нигде не доказано отсутствие других решений, переход к системе ничего не обосновывает.
Many Many Thanks to you. ❤❤❤
This is so overly complicated.. Just check low values of m and find that m = 4 works.. Then use a simple analysis tool to show uniqueness of the solution e.g. by showing that function f(m) = 3^m - 2^m is strictly increasing for m>=1.
Additionally your solution contains errors and missing steps:
1. If you do the trick with 3^(m/2)^2 - 2^(m/2)^2 = 65 then you presuppose that m is even, because if m is odd then 3^(m/2) is not natural anymore, so you cannot use the natural divisors of 65 in the following steps anymore.
2. Even if that approach worked (because you somehow proved that m must be even): After you rewrite the equation as x^2 - y^2=65, then you have to consider *two* pairs of solutions 65 = 65 * 1 and 65 = 5 * 13.
Long story short, lots of mistakes in your video unfortunately.
Thanks for this keen observation and I really appreciate this comment sir.
Noted. I will do more detailed work on subsequent videos.
You the best and much love for this detailed comment....💕💕💕
Yes this video is shit full of errors. The fact that it has so many errors and is hugely overcomplicated at the same time makes it complete and utter shit
3ᵐ-2ᵐ=65
y=3ˣ-2ˣ
y=65
если построить оба эти графика, то будет видно, что уравнение имеет единственное решение, поэтому можно попробовать подобрать корень подбором:
х=1: 3¹-2¹=1; 1≠65
х=2; 3²-2²=5; 5≠65
х=3; 3³-2³=19; 19≠65
х=4; 3⁴-2⁴=65; 65=65
х=4 корень
m=4
Chính xác.anh biến đổi 65=5×13.đây mới là mấu chốt của bài.và tư duy hàm số mũ...tuyệt.
in maths olympiad time is very critical...what is important is the answer...so the best way is to solve maths olympiad is to have very basic maths then the rest is analysing to get the pattern...so what i did is just look at indecies of 3 that have the last number such that if we subract an indecies of 2 and get 65
Excellent explanation. It seemed very complicated, but it turned out to be easier than expected. A hug
Excellent ! Solution becomes so easy and understandable
if m0 and the derivative of 3^m-2^m is 3^m*log(3) - 2^m*log(2) which in turn larger than 3^m*log(2)-2^m*log(2) which is clearly positive as both log(2) > 0 and 3^m>2^m hold. Therefore the solution is unique, since 3^m-2^m-65 is increasing. A simple guess shows m=4 is a solution.
Your work opens up the horizon of my mind how to approach this type of question. Thank you.
Thank you very much!
Без перевода мне всë так понятно было👍
Criticizers will be always there just ignore them.Good work brother👍
The presentation and demonstrated solution is very good. rubikaz comment is nevertheless valuable. If the approach solves, then the uniqueness provides all solutions. Using an exclusive deduction from the start is valid, if a working hypothesis. Thanks for this video.
I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚
1. (3^m-2^m) isincreasing function of m.
2.M is integer,because 65 is integer.
3.f(0)=0,f(5)=243-32=211.
4.we only need to test by 1,2,3,4
5.The answer is 4
Got something helpful from the procedure.
amazing explaination of procedure
You worked hard, I found the result by giving the value of m to 4 in 10 seconds, but it is important how the solution is, but I still think it is too long. Greetings from Turkey.
Thank you very much my dear friend
Difference of squares into u-substitution. Excellent methodolgy.
Thanks a bunch sir.
Beautifully explained.
So helpful . And this video gave me idea how to solve elzebric brobepems ❤ .
Take more video related to solve many brobepems in polynomial . Logorithimc function and trigonometry
Nice and very helpful video
Nice one ...... good teacher ...... stay teaching stay doing more math tricks ...... i liked it
Mathematics is all abt observation and understanding thee pattern thx a ton Buddy❤
Thanks a million for watching and dropping this wonderfully encouraging comment sir. The true is that mathematics is all about finding a solutions to problems/challenges. We really appreciate the fact that you watch and dropped this comment to clarify some debts in the minds of so many viewers and subscribers here.
Much love from all of us @OnlinemathsTV....💕💕💕💖💖💖❤️❤️❤️
Wonderfully explained, very informative.
But it’s sometimes more convenient to use the easy method
By the trial and error method
It easy to solve....
Sir you have understood us fine way
Thank you. It is very usefull for me.
Elegant solution 👏🏻👍🏻
As a math teacher, this is a plus to me. You are amazing 👏
Thanks a million sir.
Basit düşünecek olursak 3^m ve 2^m degerleri fonksiyon oldugundan fonsiyonlarda m icin koordinat düzlemi mantigiyla hareket edersek bu tip sorularda yapilacak en kestirme yol m degerleri yerine m=0,1,2,3,4,5,... Ve negatif taraflardan -1,-2,-3,-4,-5,... Degerlerinden verilince direkt cevap cikar. M= 4 direkt cevap cikar. Birkac denemeden sonra direkt cevap çıkması olasıdır. (Fonksiyonda noktayi yazip sonucu bulmak gibi). Bircok soruda direkt cevap cikar. (Bircok sorunun cevabı da iki kare farkı, toplamın karesi, farkın karesinden direkt cevap çıkar.). Bircok olimpiyat sorusu da bu metotla çözülebilir. Tamami degil ama sorularin çözümlerinde saglam, kısa süreli çözersiniz. Öğretmenin emegine saglik, cok güzel anlatmış bu arada
Nice , solution & was very enlightening. Thanks
I just use hit and trial method and easily get ans within 2-3 sec in thumbnail but this video concept is really good
Gentleman I appreciate your work.
Wonderful tutor.Keep it up sir.
m = 4. It's the lowest integer that makes 3^m bigger than 65, so it's the first one to check; this can be done in one's head.
Nice solution. It makes so easy, the way you explained.
Great lesson! Thanks. (From São Paulo/BR)😃
Excellent explanation
Amazing work.. Thank you for sharing...
This maths was wonderful I learnt a whole lot from it
We are glad you learnt something from this sir.
Wonderfully done
3^m - 2^m = 65
2^m{(3/2)^m - 1}=65
2^m{(3/2)^m - 1}= 2^4{(3/2)^4 - 1}
Thus m=4
Before watching this video i want to share my method -
By using binomial expansion
3^m - 2^m =65
=>(1+2)^m - (1+1)^m =65
=>(1+2m + (2m(2m+1)/2!) +........) - (1+m+(m(m+1)/2!)+.......)= 65 (by binomial expansion)
Bro i got stuck 😮💨
You are a genius 🎉🎉
You are a genius 🎉🎉
I was able to solve the problem in less than 1 minute... I just thought about the Multiple of 3 more than 65 that was 81 and then i subtracted 65 from it and the resulting answer was 16,now we know that 3^4 is 81 and 2^4 is 16,so easily the value of m came out to be 4
.
I have made
3exp1 -2exp1=1
3exp2-2 exp2 =5
3exp3-2exp3=19
3exp4-2 exp4=65 81 -16 = 65
Différence between two perfect squares
It was wonderful! Don’t show the cancellation of base . One has to argue that when bases are same, indices are same. So (m/2) = 2
thank you sir