A way to solve the problem is to recognize that 3^m has to be greater than 65. Therefore, m must be greater than 3. Then starting with m=4 you get the solution immediately.
WHY must m be greater than 3? Only because you have to first work out 3 to the power of 3, which is 27. In fact you have to first work out 3 squared, which is 9. So you're not starting at 3 to the power of 4, you're starting at 3 squared. Yes, i know that we all know what 3 squared is automatically. But you're assuming we all know that 3 cubed is 27 instantly, which we probably do. Well what if we also know what 3 to the power of 4 instantly is? We all have different levels of maths skills and/or memory, so we all need to start at different values of m (probably 3 or 4), not always just 4...
@@stevespenceroz It is obvious that 3^m must be greater than 65 since we subtract 2^m to get 65. it is also obvious that m must be >3 since 3*3*3 = 27. It is also obvious that 3^4 is greater than 65 since 3^4 is the same as 3^2*3^2 = 9*9 = 9^2 = 81, which is greater than 65. Then starting with m=4, 2^4= 2^2*2^2 = 4*4 = 4^2 = 16. Obviously 81 - 16 = 65. QED
Geez, after all of that algebra you ended up using trial and error to get 64 and 81. That could be done right at the very beginning, without even having to write anything down! You did that first in the video, and it was pretty simple. Why go through all that other rigamarole? I thought you were going to substitute 3^(m/2) and 2^(m/2) for a and b, respectively, but I see that that just gets us back to where we started.
Indeed the whole video is ridiculously uninformative. It can't even be considered a math problem with a solution being guessed. Even math videos have their spam now. Glory to the few cents the poster will get out of tricking us to his garbage posts.
My thoughts exactly. He just moved the trial and error method to later, lol. Also, he spent like 3 minutes explaining the a^2-b^2=(a+b)(a-b) identity, then promptly ignored it and just started plugging guesses into a^2-b^2. Waste of 22 minutes. Now I'm wondering if there really is a way to solve this algebraically.
I would suggest another way to find out the solution. After arriving to (a+b)(a-b)=65, we noticed that 65 can only be obtained multiplying 65 by 1 or 13 by 5. If we assume a+b=13 and a-b=5, after adding both equations we have 2 x a = 18, then a=9. If 3 powered to m/2 iquals to 9 (or 3 powered to 2) then m/2= 2 and m=4.
We can solve it in another way from the concept of a² - b² = (a+b)(a-b). For example, we we have already written(3^m/2)² - (2^m/2)² = 65. Since this is in the form of a² - b² = (a+b)(a-b), we can write this as (3^m/2 + (2^m/2)(3^m/2 - 2^m/2) = 65. Considering m as a positive integer, the LHS is the product of two terms. Also (3^m/2 + (2^m/2) > (3^m/2 - 2^m/2). Again 65 is the product of (i) 13 and 5 and (ii) 65 and 1. There is no third pair other than these two. Therefore, we two cases, (a) (3^m/2 + (2^m/2) = 65 and 3^m/2 - 2^m/2) = 1 and (b) (3^m/2 + (2^m/2) = 13 and (3^m/2 - 2^m/2) = 5. Solving the first set (a), that, adding them, we get 2. (3^m/2) = 65+1=66, that is 3^m/2 = 33. We cannot get a solution for this, since no power of 3 is 33. Solving the second set (b) in the same way, we get 2. (3^m/2) = 13+5=18, which gives 3^m/2 = 9 =3². From this we get m/2 =2, meaning m =4.
"Considering m as a positive integer"? Why? Who says you can just consider something to be something. What if it's not 65, but instead a value that means m is not an integer...?
In Europe, in the 60's from 6th grade to 12 th grade each subject was graded with numbers. Not letters. Graded out of 20. And you needed to have 10 to pass. All grades of each subject were added every month and divided by the numbers of tests. But each subject had a coefficient, meaning some were worth more than others. Like in college. Some courses get 3 credits some get 4. So if you got an 8/20 in english for instance and a 16/ 20 in history you would add the extra 6 to English and you passed. Well... For 7 years I had 1/20 in math sometimes 1.5/20. Hahaha. And I graduated with a 12/20 Then I went to college and only took one math course Algebra 1. Thank you. But I still don't get in. Lol 14:56
As I've seen in the comments that it is ridiculous to get to the algebra and start trial and error. Starting with (a-b)(a+b) = 65 --> (a-b)(a+b) = (5)(13) Since a-b should be smaller than a+b we should get a - b = 5; a + b =13 --> 2a = 18 --> a = 9; 9 + b = 13 --> b =4 therefore 3^(m/2) = 9 = 3^2 --> m/2 = 2 --> m=4 similarly with 2^(m/2) = 2
@@WillamGorsuch You're right. Luckily I didn't work out the b part in detail like I did the a part. The idea is that trial and error is unnecessary. While your point is don't rush and check your work.
To me, the easier approach was to go above the 65 with the 3m power because we will subtract 2m power back to the answer of 65. 3 to the 3rd power is only 27 and not above 65, so 3 to the 4th power is 81. 81-65=16=2m power or m=4. So it all plugs in. I don't know if I could remember how to do it in the video without seeing the entire thing on paper to follow.
Solved it in my head by Binomial expansion / Pascal's triangle: Rewrite 3^m as (2 + 1)^m and you will get the first coefficient (which is always one) canceled out because that's always 1 • 2^m and we subtract that at the end. To shorten the calculation, remove 1 from the result (65 -> 64), so the 1^m, because it will always lead to 2⁰ at the end. Check for m = 3: 3 • 2² + 3 • 2 = 12 + 6 = 18 too low Check for m = 4: 4 • 2³ + 6 • 2² + 4 • 2¹ = 32 + 24 + 8 = 64 => m = 4.
This whole trick (substituting with a^2 - b^2) is also guessing ("let b be 1, oh no that is not possible, so let b be 4, then b^2 = 16" and a^2 of course 81 (81 - 16 = 65) and so a is sqrt. 81 = +/- 9 (here of course + 9). In fact it is a more fancy guessing method. :)
You seem to forget that guess and check is a valid math method. In the expression we see that 3 and 2 must have the same exponent to produce the required result of 65. m=3 produces 3x3x3 = 27 minus 2x2x2 = 8 . 27-8 = 19 . m= 4 Produces 81 - 16 = 65 . On the second try.
Once you think 3^3 = 27, you can stop right there - you're going to subtract from that and it's not enough. You don't need to do the - 2^3. m=4 is the first one that's even worth working all the way.
It seems that the quest for a solution at 18:43 reverted essentially to using the same trial-and-error as used at 3:28. Clearly, a more elegant way at 18:43 is to use the products method, whereby 65 is factored into 13 and 5; with the proviso that (a + b) > (a - b). The expression yield two simple equations that are solved simultaneously to derive the values of a and b, respectively. Thus, (a + b)(a - b) = (13)(5). From which, (a + b) = 13 ..........(1) (a - b) = 5 ..........(2) Solving simultaneously (1) and (2) gives 2a = 18 a = 9. From the earlier substitutions, a = 3^(m/2) That is, 9 = 3^(m/2) or 3^2 = 3^(m/2) Thus, 2 = m/2 or m = 4.
Difference of 2 squares and splitting 65 into its only 2 factors. Integers only. (a+b)(a-b)=13 x 5 a+b=13, a-b=5 adding and subtracting these 2 eqns we get 2a = 18, 2b = 8 a=9, b=4 3**(m/2)=9=3**2 (m/2)=2 m=4 ** was the old original basic, algol, fortran programming syntax to exponentiate.
Hi John, I enjoy your lessons. In this lesson on your second step I noticed that the answer is right there. Three times the exponent denominator and two times the exponent denominator. That is 3 ^2 and 2^2. Is this a coincidence, or is it a consequence of exponent properties?
Hey there, Mr. TH-cam Math Man! I've been really intrigued by this problem! Please note that for the duration of this post, I'm using x and y in place of a and b. As far as I can tell, you can only solve for the value of m using the difference of squares formula, x^2-y^2=(x+y)(x-y), when m is both a product of 2 and is greater than or equal to 2. In other words, an even number greater than or equal to 2. Thus, m can have a value of 2, 4, 6, 8, 10, 12, and so on. This is a direct result of the conditions established at the beginning of the problem by making x=3^(m/2) and y=2^(m/2). Therefore, m itself must always be a product of 2 in order to yield a whole number. For example, as seen in this problem, if m=4, then m/2=4/2=2. However, if m=3, you end up with the ratio m/2=3/2, which obviously doesn't yield a whole number when divided. Anyway, the value on the right side of the equation will need to change depending on the value of m so that the equation remains true. If m=4, we get 3^m-2^m=65, the equation used here. If m=2, the equation becomes 3^m-2^m=5. However, if m has a value that isn't a product of 2, you can't use the difference of squares formula to solve for it. Regardless of which pair of factors you use, you will end up with two numbers that can't be broken down into an exponent of either base 2 or 3. Even if you try using logarithms, you won't obtain the correct value for m. For example, when m=3, the equation becomes 3^m-2^m=19, which can't be solved using the difference of squares formula. However, it can be solved using the difference of cubes formula, where x^3-y^3=(x-y)(x^2+xy+y^2). In this case, the equation takes the form of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3). However, just as using the difference of squares formula requires that m be a product of 2, so too does using the difference of cubes formula requires that m be a product of 3. In other words, m is both a product of 3 and is greater than or equal to 3. Thus, m can have a value of 3, 6, 9, 12, 15, 18, and so on. Once again, this is a direct result of the conditions established at the beginning of the problem by making x=3^(m/3) and y=2^(m/3). Therefore, m itself must always be a product of 3 in order to yield a whole number. For example, if m=3, then m/3=3/3=1. However, if m=4, you end up with the ratio m/3=4/3, which obviously doesn't yield a whole number when divided. Interestingly enough, just as using the difference of squares formula gives one set of (x,y) values, using the difference of cubes formula gives two sets of (x,y) values. In the case of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3), one set of (x,y) values will be positive while the other will be negative. Obviously, only the positive set of (x,y) values will yield the correct solution. If m=6, the equation becomes 3^m-2^m=665. In this case, the value of m can be obtained by using either the difference of squares formula or the difference of cubes formula because 6 is evenly divisible by both 2 and 3. m/2=6/2=3 and m/3=6/3=2. For this reason, both formulas can be used when m has a value of 6, 12, 18, 24, 30, 36, and so on. This further implies that if m=4, you could also use x^4-y^4. If m=5, you would use x^5-y^5. If m=6, you could also use x^6-y^6, and so on. Obviously, using this approach will make your equations far more complex and not very practical each time the value of m increases. This also implies that using the difference of either squares or cubes won't solve for values of m which aren't evenly divisible by either 2 or 3, such as 5, 7, 11, 13, 17, 19, and so on. Also, things get kind of weird when m equals either 0 or 1. To sum up, use the difference of squares formula when m is a product of 2. Use the difference of cubes formula when m is a product of 3. And use either formula when m is a product of 6. Anyway, getting back to the original problem, when m is a product of 2, is greater than or equal to 2, and thus can be solved for using the difference of squares, only the middle pair of factors of the number on the right side of the equation will yield the correct solution. For example, if we take the equation presented in the video, 65 has four factors, which are 1, 5, 13, and 65. Only by using the two middle factors, 5 and 13, do we obtain the correct solution. Similarly, when m=8, the equation becomes 3^m-2^m=6305. 6305 has eight factors, which are 1, 5, 13, 65, 97, 485, 1261, and 6305. Once again, only by using the two middle factors, 65 and 97, do we obtain the correct solution. However, when m is a factor of 3, is greater than or equal to 3, and thus can be solved for using the difference of cubes, then it's possible for another pair of factors of the number on the right side of the equation to yield the correct solution instead of the middle two. This is especially true when m has a value of either 6 or 12. In any case, it's a process of rejection and elimination to find which set of factors gives the correct value of m. Finding the factors of any number is easy when using a website such as Number Empire. In addition to algebra, graphing is another ideal method for finding the value of m. If we use the equation presented here, 3^m-2^m=65, let m=x, then move everything to one side and make it equal to 0 or y, we get y=3^x-2^x-65. This produces a logarithmic graph with (x,y) intercepts at (4,0) and (0,-65). In other words, the graph intersects the y axis at -65 and the x axis at 4. If x=m, then 4 is our answer. For equations that prove difficult to solve using algebra, such as when m equals 0 or 1, equations involving the sum of squares, such as 2^m+3^m=13 when m=2, or when m is a value indivisible by either 2 or 3, such as when m equals 5, 7, or 11, then graphing is a viable, convenient, and useful alternative to find the value of m. Anyway, I would love to hear your thoughts on this! Or just graph it for fun! Either way, thanks so much for the content you post here on TH-cam! I find your discussions really entertaining! Keep up the good work, Mr. TH-cam Math Man! Grá ó Colwyn 🌝🦉🌲
Difference of two squares is a good method. However, one must notice that 65 is the product of two primes. The prime numbers are 5 and 13 A+B = 13 A-B = 5 Solving this yields A = 9, and B = 4 However, plug and guess got me the answer in much less time than the more rigorous solution using prime factorization.
But isn't there an analytic solution to the general case of B1^x - B2^x = C ? Generally measured numbers won't be exact integers, let alone factorable ones. Lets say for ex. these numbers show up when calibrating some sensors with an exponential response. And we'd require the exact exponent that matches the measured values.
Yes, I would like to know if there is an algebraic technique for solving this general equation that doesn't involve guessing, graphing, or doing a bunch of algebraic manipulation and guessing again. I expect in this case there is no way to isolate m.
For m < 0, there can be no solution, since the absolute value of 3^m - 2^m is less than 2, or even 1. For m > 0, 3^m - 2^m is strictly increasing (just differentiate it). Thus if one guesses a single solution, it must be the only one. Always try guessing small integers just to test, and one see m = 4 works.
Interesting series for a^m - b^m = x, where a>b and m is an integer. For power of 2, nested squares: x = 2b(a-b)+(a-b)^2 For power of 3, nested cubes: x = (2b(a-b)+(a-b)^2) *b+a^2 For power of 4, nested 4-cubes: x = ((2b(a-b)+(a-b)^2)*b+a^2) *b+a^3 For power of 5, nested 5-cubes: x = (((2b(a-b)+(a-b)^2)*b+a^2)*b+a^3) *b+a^4 Each higher power tacks *b+a^(power-1) to the end of the previous level's equation. Spiffy. My eyes are crossing after that, but I think I got it right.
It doesn't give you the full solution, but it is not so hard to prove that m must be even. Goes like this : split off the highest power of 2 from 65. This gives you 65 = 64 + 1 = 2^6 + 1. Now substitute that, and bring the 1 and 2^m to the opposite sides, and you get 3^m -1 = 2^6 + 2^m. Now you can split off a power of two in the right side. Concerning only whole numbers, there are two possibilities : 2^6(2^(m-6) + 1) for m >= 6 and 2^m(2^(6-m) + 1) for m
I once took an algebra test in high school which had to do with Johnny selling some of his newspapers at X price per paper and later in the day some at Y - find Y. I used trial and error and found 4 cents. (It was a long time ago!) The next day I got the test back with the question marked wrong, even though 4 was right, because I did not use algebra and show my work. The teacher said grumpily, "Albert, I was going to accuse you of cheating, but nobody else got the answer." I left a trail of D's in math over the next 5 years, but pursued and completed my doctorate - (in music!) If you need to know math learn the formulas and equations. I didn't need to!
1.- Let be a=3^x ; b=2^x; 2.- this means x=log10(a)/log10(3); but also x=log10(b)/log10(2); log change base general formula : logn(s)=logm(s)/logm(n); 3.- System of 2 equations with 2 unknowns, 1 eq is non-linear : eq1: a-b = 65 ; eq2: log(a)/log(b)=log(3)/log(2) = k1; log same as log10 4.- putting eq2 into eq1 : b^k1-b = 65 With my wonderful HP48 or solving graphically : b = 16; 5.- if b=16; then x=4; 6.- check 3^4-2^4=65 ok
I loved the general method, it gives a scale independent solution methodology. The problem as stated was solved instinctively by many, myself included, through learning the times table in primary school.
3^m - 2^m = (m = 2k) = 3^2k - 2^2k = (3^k - 2^k)(3^k + 2^k) (by difference of squares). An obvious factorization of 65 is 5 * 13, and the sum is larger than the difference. Let the difference 3^k - 2^k be equal to 5. This is easily solved by k = 2, so m = 4. Sanity check: verify that the sum is 13, ie. 3^2 + 2^2 = 13. Second test: verify that (3^4 = 81) - ( 2^4 = 16) = 65. You could use modular arithmetic to rule out other candidates.
Well,, the other approach is 3^m - 2^m = 65 3^m(1-(2/3)^m) = 65 ln( 3^m(1-(2/3)^m)) = ln(65) m ln(3) + ln(1-(2/3^m)) = ln(65) m = ln(65)/ln(3) - ln(1 - (2/3)^m) / ln(3) since 1 - (2/3)^m is going to be between 0 and 1, ln(1 - (2/3)^m) / ln(3) is going to be a small negative value, and thus our answer will be slightly more than ln(65)/ln(3) which is 3.799, so we can conclude it's 4.
This particular problem involves a combination of various algebraic concepts and strategies such as exponential laws, distribution laws, substitution techniques and factoring strategies. Although it's easy to guess the answer is 4, many students would indeed have trouble solving it algebraically . It may seem simple at first glance but it's easily one of the most difficult problems that TabletClass Math has published.
I did roughly the the same as you and ended up with (sqrt(3)^m + sqrt(2)^m) x (sqrt(3)^m - sqrt(2)^m) = 13 x 5. (Since 65 = 13 x 5.) Then I let sqrt(3)^m + sqrt(2)^m =13 and sqrt(3)^m - sqrt(2)^m = 5. Adding those two equations together gives 2xsqrt(3)^m = 18. Simplifying and squaring both sides yields 3^m = 81. Thus we know m is 4.
4 3^m = 65 + 2^m Hence 3^m is at least 81 or 3^4 since m is an integer and since 65 is greater than 3^3 or 27 Hence m is at least 4 Try 4 3^4 -65 = 2^4 16 = 2^4
Is the follwing method of fining answer acceptable to the Board? I resoned that 3^m has to be greater than 65 ...so first such power is 3^4 =81....I tried .... "81-2^4" is eactly 65!!!...... m =4.... IS THIS considered legal for high score?
It was not assumed that m is an integer. Even if that is the case, there is no reason for a=3^(m/2) and b=2^(m/2) to be integers. The only reasonable way to do this problem is to notice that for positive x, the function f(x)=3^x-2^x is increasing (hence injective) and therefore the solution will be unique - then we can get the solution by trial and error, or estimate the solution in case the answer is not nice (e.g., f(x)=64 also has a unique solution, but the answer is not an integer ). Then only interesting part of the problem would be to ask: where is the function f defined above is increasing? (Ans. for x>= -ln(ln(3)/ln(2))/ln(3/2) approximately -1.14).
With trial and error, I can get an answer, m=4. I of course still have to prove that this is the only solution I can first say that, if m 0, and I can take the derivative wrt m of 3^m-2^m, giving me 3^m ln3 - 2^m ln2. Since ln3 >ln2 >0, 3^m ln3 > 2^m ln2, so the LHS is monotonically increasing wrt m. Therefore, m=4 is the only solution.
I was hoping to see an algebraic solution as well (if one exists) and found the trial and error solutions pretty disappointing. However, given that's what we have, couldn't we prove m=4 is a unique solution by simply noting that decreasing m always makes the right side smaller and increasing m always makes the right side larger? I'm sure there's a way to formalize that better, but hopefully, you get what I mean.
@@jeffdege4786 For some reason YT suggested this video and I don't really want to spend time on the prolix solution presented in the video. Showing 3^x-2^x is monotonic increasing is a key step, but how to justify it without calculus? Perhaps start with the observation that 3^x is just 2^x but "squished" in the x-direction.
@@peterbrockway5990 3^x - 2^x = 2^x (1.5^x - 1). This is a product of two functions which are both monotonically increasing and which are both positive for x > 0. Which implies that the product itself also is monotonically increasing.
The trouble with your “creative” method shown at ~15:00 is that by taking the square roots, you may end up with a and/or b which are not integers, even if m was. Then, you’ve no reason to assume that their sum and/or difference is an integer and therefore, the factorization of the right side is wrong as well. Indeed, if you try this with 3^m - 2^m = 19, you’ll end up with only one way to factor 19 as it is a prime: 19 x 1. This gives you a = 10 and b = 9. So now m should be log10/log(sqrt 3) as well as log9/log(sqrt 2)… Which isn’t even equal, let alone correct (as m = 3). I know you’re trying to show some less rigid, analytical methods, but as soon as you make any such assumption, you absolutely must note it, with a great exclamation mark and always check any result you obtain like that… I didn’t hear anything.
I just used guess and check. I knew it had to be a fairly small exponent because different bases would diverge very quickly, the difference is already pretty large at the exponent of 6.
So either trial and error the indices to start with, or trial and error a difference of squares?! If it's trial and error the first is easier. The difference of squares route shows working but it's not a proof because... it's trial and error. Or am i wrong?
I found by it by trial and error. First wanted to use small numbers to see if i could see a principle at play. Used 3, which is 27 - 8 = 19. So then I hoped 4 would be the right answer, and it was. But, I have not finished the video yet, so I hope there is a slicker way to solve this. Thanks.
@@wrc1210 Yes and No. Granted it was a very long way to get around the guess work...so in that sense, not very practical for me, anyway, but I did find it interesting...that it could be done. Did you watch to the end? Sometimes trial and error is the most efficient method in real life. That, in and of itself, perhaps is the most valuable lesson here.
@its-a-bountiful-life But he didn't "get around the guess work." He just moved it to a different spot after manipulating the equation a bunch and making it harder to do the guess work than it was in the original form. I don't know. Sure, he demonstrated some algebra that might be useful in solving other equations, but it wasn't at all helpful in this one. Not a big deal, but kind of annoying. Wouldn't care as much if he was up front at the beginning and just said I'm not going to show you an algebraic solution to this, but here are some dos and don'ts about manipulating these types of equations and you'll see why this is such a difficult problem to solve without resorting to guess work.
Could you please show how to solve this if the sum was 67 instead of 65? I’d like to see the solution for that one which didn’t depend upon guessology: 3^m - 2^m = 67 Solve for m
Thanks for that. I originally started down all sorts of dead ends, then found the simple trial-and-error of method 1. However, I really enjoyed the last technique, even though that also involved some T&E.
Plot 3^m and 2^m. The difference as m increases is observed to increase. There is therefore only one real solution. Simple inspection reveals m=4. There may be imaginary solutions.
at 17:49 we have to add a value to make the result a "better number". Why has that better number to be a perfect square? a=2^(m/2). May be the original equation has also another solution, an odd value maybe. Then a is not a perfect square. So it is not correct to use the fact that a has to be a perfect square, as you do in your video.
Maybe I did too much math ;-.) .. But I "see" .. 81 - 16 almost instantly (meaning to the power of 4 for both numbers) ^^^ But going the algebraic way .. at the point where you have (a+b)(a-b)=65 .. this calls for a product of 2 prime factors 13 and 5 ... which yields 9 and 4 pretty quickly which gives m=4 in return
I tried doing it a different way using logs and I got 2 different answers 3.700 and 4.41. What am I doing wrong? Divide the terms on the LHS by 2^m If you do 2^m([3/2]^m - 1)= 13 x 65. (Just to check, 2^m x (3/2)^m = 3^m) 2^m = 13; mlog2 = log13; m= log13/log2 and I get 3.700 (3/2)^m -1 = 5; (3/2)^m = 5 +1; (3/2)^m = 6; mlog1.5 = 6; m = log6/log1.5; m = 4.41. Please let me know where I have gone wrong😂. My maths has gone
3^m - 2^m = 65 I remember that 81 - 16 = 65, so I can rewrite them as powers of 2 and 3: 3^4 - 2^4 = 65 Thus, m = 4 . There's no need to step through integers 0, 1, 2, 3, ..., for 3^m - 2^m - 65 = 0 unless there's no integer solution. In that case, we can find where the potential m value is between two successive integers that straddle 0, if at all. If no crossovers are found, then there's no solution.
I knew the factors of 65 as 5 & 13, and if I used the difference of squares, the factors are the difference and sum of 3^(m/2) and 2^(m/2). The average of the factors is 3^(m/2) and 9, so m = 4. As a check, half the difference is 2^(m/2) and 4, which also has m = 4.
If you get to (x+y)(x-y)=65 then factor to (x+y)(x-y)=13x5 then break it out x+y=13 and x-y=5, you can solve for each and find that x=9 and y=4, and you're there.
If m is negative then 3^m and 2^m are in [0, 1] and their difference cannot be equal to 65, so if the given equation has solutions they are positive. f: R+ ---> R+, x ---> 3^x - 2^x strictly increases on R+ (positive derivative), so if the given equation has a positive solution then it is unique. As x = 4 is evident solution, it is the only positive one and finally the only real solution.
I once had to solve a fluid mechanics problem in an exam, where one first had to derive the formula for the flow rate through a sluice gate and then solve the formula to obtain the flow rate. Once derived, the equation for the flow rate could not be solved by algebra and most students gave up and moved on to the next question, thinking that the formula they had derived must be wrong. After the exam, the professor explained that there was no need to use algebra to obtain the answer and that a numerical solution would have been acceptable. In this case, I would set the equation to zero, (3^m-2^m)-65=0. Then I would find a trial value for m that gives a positive answer to the equation and a trial value for m that gives me a negative answer. I would then find the root of the equation for zero by using a numerical algorithm such as the bisection method for example. In this case, it was not necessary for me to do this because I found the root of the equation while I was searching for two suitable trial values of m. If the answer was not a whole number, it would have been different.
I'm not happy with trial and error approach used here. Did some searching and this seems to be solvable by using the Lambert W Function. ChatGPT almost shows how to do it. Will the future math classes be taught by AI?
Good golly - so many words! Like those web pages that draw you on and on to keep you scrolling through all their adverts, to the trivial end. Never again. Unless I need something to put me to sleep.
@@danv2888You're correct. I don't. That's why I just switched to gender studies. My first assignment is defining what a woman is. I can solve that one.
As the narrator stated in the video, the equation does NOT lend itself to solution by the rules of logarithms. It *looks* like it does, but it does not. You can't simplify log (3^m - 2^m)
I would solve it using modulos. First mod 3 shows m even: m = 2a. 9^a - 4^a = 65 mod 8 gives a even: a = 2b. Then: 81^b - 16^b you can always factorize: 81^b - 16^b = (81-16) (81^(b-1) + 81^(b-2)*16 + ... + 16^(b-1)) This sum is always a positiv integer. Therefor 81^b - 16^b >= 81-16 = 65. And equality only if the above sum is 1, which is the case only for b = 1, and thus n = 4.
It's easy to "cheat" and get the right answer in a few seconds by trying, "1,2,3,4" in your head, but if the answer was not a simple integer, the next level of "cheating" would be to observe that 3^m grows faster than 2^m, so the function will grow monotomically with "m", and you can write a computer script to do a binary search, That may still be "cheating", but practicing with such scripts can be a valuable skill for its own sake, extendable to other kinds of problems that don't lend themselves to other methods. But still spend the 22 minutes to learn the algebraic methods, as well.
When author got a^2-b^2=65, after this he overcomplicated. I will recommend (a-b)*(a+b)=5*13; Therefore, a-b=5 and a+b=13, 2*a=18: a=9: b=4; When a-b = 13 and a+b=5: 2*a=18; a=9; b=-4; If (a-b)*(a+b)=1*65; then a-b=1, a+b=65 or a-b=65 and a+b=1. When we solve a=33, b=32 and a=33; b=-32.
Not sure about the algebra without some head scratching, but got the answer in about 10 secs with trial and error. I started with M = 3 and it didn’t work. (3x3x3)= 27 - (2x2x2) = 8 so wrong answer of 19 Then M=4 it worked 3x3x3x3 = 81 2x2x2x2 = 16 82-16= 65
I did this one thanks to a lucky guess to be quite honest. m had to be 'more than 3' because 3^m had to be bigger than 65 and 4 came next .. 3^4 - 2^4 = 81 - 16 = 65
A way to solve the problem is to recognize that 3^m has to be greater than 65. Therefore, m must be greater than 3. Then starting with m=4 you get the solution immediately.
👍 Exactamundo!
WHY must m be greater than 3? Only because you have to first work out 3 to the power of 3, which is 27. In fact you have to first work out 3 squared, which is 9. So you're not starting at 3 to the power of 4, you're starting at 3 squared. Yes, i know that we all know what 3 squared is automatically. But you're assuming we all know that 3 cubed is 27 instantly, which we probably do. Well what if we also know what 3 to the power of 4 instantly is? We all have different levels of maths skills and/or memory, so we all need to start at different values of m (probably 3 or 4), not always just 4...
@@stevespenceroz It is obvious that 3^m must be greater than 65 since we subtract 2^m to get 65. it is also obvious that m must be >3 since 3*3*3 = 27. It is also obvious that 3^4 is greater than 65 since 3^4 is the same as 3^2*3^2 = 9*9 = 9^2 = 81, which is greater than 65. Then starting with m=4, 2^4= 2^2*2^2 = 4*4 = 4^2 = 16. Obviously 81 - 16 = 65. QED
No idea how I worked it out. It felt like the answer was 4. Checked it, and it was. But don't know how I did it
Qaqaaq@@jim2376
Geez, after all of that algebra you ended up using trial and error to get 64 and 81. That could be done right at the very beginning, without even having to write anything down! You did that first in the video, and it was pretty simple. Why go through all that other rigamarole?
I thought you were going to substitute 3^(m/2) and 2^(m/2) for a and b, respectively, but I see that that just gets us back to where we started.
I solved it in my head using trial and error in seconds. How to solve it "correctly," I'm not sure.
Indeed the whole video is ridiculously uninformative. It can't even be considered a math problem with a solution being guessed. Even math videos have their spam now. Glory to the few cents the poster will get out of tricking us to his garbage posts.
My thoughts exactly. He just moved the trial and error method to later, lol. Also, he spent like 3 minutes explaining the a^2-b^2=(a+b)(a-b) identity, then promptly ignored it and just started plugging guesses into a^2-b^2. Waste of 22 minutes.
Now I'm wondering if there really is a way to solve this algebraically.
Solving this algebraically would need logarithms, at which my brain just goes "no" and walks away.
Trial and error worked for this case.
@@AlainPaulikevitchI agree. This video is ridiculous, once it gets passed the first solution. Just a big waste of time. Some might call it stupid.
I was keeping up until about @16:20, when it looked to me like he just decided to pick random numbers for b.
I would suggest another way to find out the solution.
After arriving to (a+b)(a-b)=65, we noticed that 65 can only be obtained multiplying 65 by 1 or 13 by 5.
If we assume a+b=13 and a-b=5, after adding both equations we have 2 x a = 18, then a=9.
If 3 powered to m/2 iquals to 9 (or 3 powered to 2) then m/2= 2 and m=4.
We can solve it in another way from the concept of a² - b² = (a+b)(a-b). For example, we we have already written(3^m/2)² - (2^m/2)² = 65. Since this is in the form of a² - b² = (a+b)(a-b), we can write this as (3^m/2 + (2^m/2)(3^m/2 - 2^m/2) = 65. Considering m as a positive integer, the LHS is the product of two terms. Also (3^m/2 + (2^m/2) > (3^m/2 - 2^m/2). Again 65 is the product of (i) 13 and 5 and (ii) 65 and 1. There is no third pair other than these two. Therefore, we two cases, (a) (3^m/2 + (2^m/2) = 65 and 3^m/2 - 2^m/2) = 1 and (b) (3^m/2 + (2^m/2) = 13 and (3^m/2 - 2^m/2) = 5. Solving the first set (a), that, adding them, we get 2. (3^m/2) = 65+1=66, that is 3^m/2 = 33. We cannot get a solution for this, since no power of 3 is 33. Solving the second set (b) in the same way, we get 2. (3^m/2) = 13+5=18, which gives 3^m/2 = 9 =3². From this we get m/2 =2, meaning m =4.
"Considering m as a positive integer"? Why? Who says you can just consider something to be something. What if it's not 65, but instead a value that means m is not an integer...?
In Europe, in the 60's from 6th grade to 12 th grade each subject was graded with numbers. Not letters. Graded out of 20. And you needed to have 10 to pass. All grades of each subject were added every month and divided by the numbers of tests. But each subject had a coefficient, meaning some were worth more than others. Like in college. Some courses get 3 credits some get 4. So if you got an 8/20 in english for instance and a 16/ 20 in history you would add the extra 6 to English and you passed. Well...
For 7 years I had 1/20 in math sometimes 1.5/20. Hahaha. And I graduated with a 12/20
Then I went to college and only took one math course Algebra 1.
Thank you. But I still don't get in. Lol 14:56
In particular, you have reduced the problem to solving two linear equations in two unknowns.
As I've seen in the comments that it is ridiculous to get to the algebra and start trial and error.
Starting with (a-b)(a+b) = 65 --> (a-b)(a+b) = (5)(13)
Since a-b should be smaller than a+b we should get
a - b = 5; a + b =13 --> 2a = 18 --> a = 9; 9 + b = 13 --> b =4
therefore 3^(m/2) = 9 = 3^2 --> m/2 = 2 --> m=4 similarly with 2^(m/2) = 2
This is a much better and logical explanation. Thanks.
@tomshane1983 you have made one mistake. 2^(m/2) does not equal 2, but 2^(m/2) equals 4.
@@WillamGorsuch You're right. Luckily I didn't work out the b part in detail like I did the a part. The idea is that trial and error is unnecessary. While your point is don't rush and check your work.
This is smart and works in this case, but you are assuming that m is an integer. What if the right-hand side was 64 or 200 or 1234.56789?
To me, the easier approach was to go above the 65 with the 3m power because we will subtract 2m power back to the answer of 65. 3 to the 3rd power is only 27 and not above 65, so 3 to the 4th power is 81. 81-65=16=2m power or m=4. So it all plugs in. I don't know if I could remember how to do it in the video without seeing the entire thing on paper to follow.
That was my method also. It did not take long and is more straightforward than the algebraic approach.
what we used to call the guessology technique
This lesson jumped to the highly extraordinary, but is a good example of why the sum of exponents can't simply distribute the exponent.
Solved it in my head by Binomial expansion / Pascal's triangle:
Rewrite 3^m as (2 + 1)^m and you will get the first coefficient (which is always one) canceled out because that's always 1 • 2^m and we subtract that at the end.
To shorten the calculation, remove 1 from the result (65 -> 64), so the 1^m, because it will always lead to 2⁰ at the end.
Check for m = 3:
3 • 2² + 3 • 2 = 12 + 6 = 18 too low
Check for m = 4:
4 • 2³ + 6 • 2² + 4 • 2¹ = 32 + 24 + 8 = 64
=> m = 4.
This whole trick (substituting with a^2 - b^2) is also guessing ("let b be 1, oh no that is not possible, so let b be 4, then b^2 = 16" and a^2 of course 81 (81 - 16 = 65) and so a is sqrt. 81 = +/- 9 (here of course + 9). In fact it is a more fancy guessing method. :)
I wish you would make another utube on the different ways of solving this equations thanks
Yes
You seem to forget that guess and check is a valid math method. In the expression we see that 3 and 2 must have the same exponent to produce the required result of 65. m=3 produces 3x3x3 = 27
minus 2x2x2 = 8 . 27-8 = 19 . m= 4 Produces 81 - 16 = 65 . On the second try.
Guess and check may work here but it's important to learn to solve these problems algebraically for situations where m could not be easily guessed.
Once you think 3^3 = 27, you can stop right there - you're going to subtract from that and it's not enough. You don't need to do the - 2^3. m=4 is the first one that's even worth working all the way.
It seems that the quest for a solution at 18:43 reverted essentially to using the same trial-and-error as used at 3:28.
Clearly, a more elegant way at 18:43 is to use the products method, whereby 65 is factored into 13 and 5; with the proviso that (a + b) > (a - b).
The expression yield two simple equations that are solved simultaneously to derive the values of a and b, respectively.
Thus,
(a + b)(a - b) = (13)(5).
From which,
(a + b) = 13 ..........(1)
(a - b) = 5 ..........(2)
Solving simultaneously (1) and (2) gives
2a = 18
a = 9.
From the earlier substitutions,
a = 3^(m/2)
That is,
9 = 3^(m/2) or 3^2 = 3^(m/2)
Thus, 2 = m/2 or m = 4.
Can you please make a video to solve 3^m - 2^m = 66?
Difference of 2 squares and splitting 65 into its only 2 factors. Integers only.
(a+b)(a-b)=13 x 5
a+b=13, a-b=5 adding and subtracting these 2 eqns we get
2a = 18, 2b = 8
a=9, b=4
3**(m/2)=9=3**2
(m/2)=2
m=4
** was the old original basic, algol, fortran programming syntax to exponentiate.
Yeah, and you get 2a=18 no matter which factor you choose to be 13 or 5, and you get a contradiction, because a,b>0, so a+b=9+b=5 is impossible.
Hi John, I enjoy your lessons. In this lesson on your second step I noticed that the answer is right there. Three times the exponent denominator and two times the exponent denominator. That is 3 ^2 and 2^2. Is this a coincidence, or is it a consequence of exponent properties?
Hey there, Mr. TH-cam Math Man! I've been really intrigued by this problem! Please note that for the duration of this post, I'm using x and y in place of a and b.
As far as I can tell, you can only solve for the value of m using the difference of squares formula, x^2-y^2=(x+y)(x-y), when m is both a product of 2 and is greater than or equal to 2. In other words, an even number greater than or equal to 2. Thus, m can have a value of 2, 4, 6, 8, 10, 12, and so on. This is a direct result of the conditions established at the beginning of the problem by making x=3^(m/2) and y=2^(m/2). Therefore, m itself must always be a product of 2 in order to yield a whole number. For example, as seen in this problem, if m=4, then m/2=4/2=2. However, if m=3, you end up with the ratio m/2=3/2, which obviously doesn't yield a whole number when divided.
Anyway, the value on the right side of the equation will need to change depending on the value of m so that the equation remains true. If m=4, we get 3^m-2^m=65, the equation used here. If m=2, the equation becomes 3^m-2^m=5. However, if m has a value that isn't a product of 2, you can't use the difference of squares formula to solve for it. Regardless of which pair of factors you use, you will end up with two numbers that can't be broken down into an exponent of either base 2 or 3. Even if you try using logarithms, you won't obtain the correct value for m. For example, when m=3, the equation becomes 3^m-2^m=19, which can't be solved using the difference of squares formula.
However, it can be solved using the difference of cubes formula, where x^3-y^3=(x-y)(x^2+xy+y^2). In this case, the equation takes the form of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3). However, just as using the difference of squares formula requires that m be a product of 2, so too does using the difference of cubes formula requires that m be a product of 3. In other words, m is both a product of 3 and is greater than or equal to 3. Thus, m can have a value of 3, 6, 9, 12, 15, 18, and so on. Once again, this is a direct result of the conditions established at the beginning of the problem by making x=3^(m/3) and y=2^(m/3). Therefore, m itself must always be a product of 3 in order to yield a whole number. For example, if m=3, then m/3=3/3=1. However, if m=4, you end up with the ratio m/3=4/3, which obviously doesn't yield a whole number when divided.
Interestingly enough, just as using the difference of squares formula gives one set of (x,y) values, using the difference of cubes formula gives two sets of (x,y) values. In the case of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3), one set of (x,y) values will be positive while the other will be negative. Obviously, only the positive set of (x,y) values will yield the correct solution.
If m=6, the equation becomes 3^m-2^m=665. In this case, the value of m can be obtained by using either the difference of squares formula or the difference of cubes formula because 6 is evenly divisible by both 2 and 3. m/2=6/2=3 and m/3=6/3=2. For this reason, both formulas can be used when m has a value of 6, 12, 18, 24, 30, 36, and so on. This further implies that if m=4, you could also use x^4-y^4. If m=5, you would use x^5-y^5. If m=6, you could also use x^6-y^6, and so on. Obviously, using this approach will make your equations far more complex and not very practical each time the value of m increases. This also implies that using the difference of either squares or cubes won't solve for values of m which aren't evenly divisible by either 2 or 3, such as 5, 7, 11, 13, 17, 19, and so on. Also, things get kind of weird when m equals either 0 or 1.
To sum up, use the difference of squares formula when m is a product of 2. Use the difference of cubes formula when m is a product of 3. And use either formula when m is a product of 6.
Anyway, getting back to the original problem, when m is a product of 2, is greater than or equal to 2, and thus can be solved for using the difference of squares, only the middle pair of factors of the number on the right side of the equation will yield the correct solution. For example, if we take the equation presented in the video, 65 has four factors, which are 1, 5, 13, and 65. Only by using the two middle factors, 5 and 13, do we obtain the correct solution. Similarly, when m=8, the equation becomes 3^m-2^m=6305. 6305 has eight factors, which are 1, 5, 13, 65, 97, 485, 1261, and 6305. Once again, only by using the two middle factors, 65 and 97, do we obtain the correct solution. However, when m is a factor of 3, is greater than or equal to 3, and thus can be solved for using the difference of cubes, then it's possible for another pair of factors of the number on the right side of the equation to yield the correct solution instead of the middle two. This is especially true when m has a value of either 6 or 12. In any case, it's a process of rejection and elimination to find which set of factors gives the correct value of m. Finding the factors of any number is easy when using a website such as Number Empire.
In addition to algebra, graphing is another ideal method for finding the value of m. If we use the equation presented here, 3^m-2^m=65, let m=x, then move everything to one side and make it equal to 0 or y, we get y=3^x-2^x-65. This produces a logarithmic graph with (x,y) intercepts at (4,0) and (0,-65). In other words, the graph intersects the y axis at -65 and the x axis at 4. If x=m, then 4 is our answer.
For equations that prove difficult to solve using algebra, such as when m equals 0 or 1, equations involving the sum of squares, such as 2^m+3^m=13 when m=2, or when m is a value indivisible by either 2 or 3, such as when m equals 5, 7, or 11, then graphing is a viable, convenient, and useful alternative to find the value of m.
Anyway, I would love to hear your thoughts on this! Or just graph it for fun! Either way, thanks so much for the content you post here on TH-cam! I find your discussions really entertaining! Keep up the good work, Mr. TH-cam Math Man!
Grá ó Colwyn 🌝🦉🌲
Difference of two squares is a good method.
However, one must notice that 65 is the product of two primes. The prime numbers are 5 and 13
A+B = 13
A-B = 5
Solving this yields A = 9, and B = 4
However, plug and guess got me the answer in much less time than the more rigorous solution using prime factorization.
In this case guess and shoot was the fastest. Mind you educated guess. m = 2 too small; m =3 gives fractional powers, start with m=4.
In both examples you simply plugged in numbers that worked.
But isn't there an analytic solution to the general case of B1^x - B2^x = C ? Generally measured numbers won't be exact integers, let alone factorable ones. Lets say for ex. these numbers show up when calibrating some sensors with an exponential response. And we'd require the exact exponent that matches the measured values.
what about 65 = 13*5 or 65*1 -> 4 cases for (a - b)(a + b)
What would be the algebraic solution for 3^m - 2^m = 2 for example?
Yes, I would like to know if there is an algebraic technique for solving this general equation that doesn't involve guessing, graphing, or doing a bunch of algebraic manipulation and guessing again. I expect in this case there is no way to isolate m.
I'm trying to solve this with logs, but it isn't working. It should work, yeah? Please help!!
Logs can't be applied to a difference of two powers with different bases in any sensible way. You first have to somehow isolate a power.
For m < 0, there can be no solution, since the absolute value of 3^m - 2^m is less than 2, or even 1. For m > 0, 3^m - 2^m is strictly increasing (just differentiate it). Thus if one guesses a single solution, it must be the only one. Always try guessing small integers just to test, and one see m = 4 works.
Interesting series for a^m - b^m = x, where a>b and m is an integer.
For power of 2, nested squares: x = 2b(a-b)+(a-b)^2
For power of 3, nested cubes: x = (2b(a-b)+(a-b)^2) *b+a^2
For power of 4, nested 4-cubes: x = ((2b(a-b)+(a-b)^2)*b+a^2) *b+a^3
For power of 5, nested 5-cubes: x = (((2b(a-b)+(a-b)^2)*b+a^2)*b+a^3) *b+a^4
Each higher power tacks *b+a^(power-1) to the end of the previous level's equation.
Spiffy. My eyes are crossing after that, but I think I got it right.
I used something like the last method. First let m = 2n, then dots. 65 = 1 x 65 or 5 x 13. 3^2 - 2^2 must equal 1 or 5. 5 works when n = 2, so m = 4
Take log on the both sides. M can be determined
It doesn't give you the full solution, but it is not so hard to prove that m must be even. Goes like this : split off the highest power of 2 from 65. This gives you 65 = 64 + 1 = 2^6 + 1. Now substitute that, and bring the 1 and 2^m to the opposite sides, and you get 3^m -1 = 2^6 + 2^m. Now you can split off a power of two in the right side. Concerning only whole numbers, there are two possibilities : 2^6(2^(m-6) + 1) for m >= 6 and 2^m(2^(6-m) + 1) for m
I once took an algebra test in high school which had to do with Johnny selling some of his newspapers at X price per paper and later in the day some at Y - find Y. I used trial and error and found 4 cents. (It was a long time ago!)
The next day I got the test back with the question marked wrong, even though 4 was right, because I did not use algebra and show my work. The teacher said grumpily, "Albert, I was going to accuse you of cheating, but nobody else got the answer."
I left a trail of D's in math over the next 5 years, but pursued and completed my doctorate - (in music!)
If you need to know math learn the formulas and equations. I didn't need to!
1.- Let be a=3^x ; b=2^x;
2.- this means
x=log10(a)/log10(3); but also
x=log10(b)/log10(2);
log change base general formula : logn(s)=logm(s)/logm(n);
3.- System of 2 equations with 2 unknowns, 1 eq is non-linear :
eq1: a-b = 65 ;
eq2: log(a)/log(b)=log(3)/log(2) = k1;
log same as log10
4.- putting eq2 into eq1 :
b^k1-b = 65
With my wonderful HP48 or solving graphically : b = 16;
5.- if b=16; then x=4;
6.- check
3^4-2^4=65 ok
Please include the points on the lines (x , y), so one can see where and how the lines crosses
I loved the general method, it gives a scale independent solution methodology. The problem as stated was solved instinctively by many, myself included, through learning the times table in primary school.
3^m - 2^m = (m = 2k) = 3^2k - 2^2k = (3^k - 2^k)(3^k + 2^k) (by difference of squares). An obvious factorization of 65 is 5 * 13, and the sum is larger than the difference. Let the difference 3^k - 2^k be equal to 5. This is easily solved by k = 2, so m = 4. Sanity check: verify that the sum is 13, ie. 3^2 + 2^2 = 13. Second test: verify that (3^4 = 81) - ( 2^4 = 16) = 65.
You could use modular arithmetic to rule out other candidates.
Well,, the other approach is
3^m - 2^m = 65
3^m(1-(2/3)^m) = 65
ln( 3^m(1-(2/3)^m)) = ln(65)
m ln(3) + ln(1-(2/3^m)) = ln(65)
m = ln(65)/ln(3) - ln(1 - (2/3)^m) / ln(3)
since 1 - (2/3)^m is going to be between 0 and 1, ln(1 - (2/3)^m) / ln(3) is going to be a small negative
value, and thus our answer will be slightly more than ln(65)/ln(3) which is 3.799, so we can conclude it's 4.
This particular problem involves a combination of various algebraic concepts and strategies such as exponential laws, distribution laws, substitution techniques and factoring strategies. Although it's easy to guess the answer is 4, many students would indeed have trouble solving it algebraically . It may seem simple at first glance but it's easily one of the most difficult problems that TabletClass Math has published.
Well if you could elaborate more on how to solve this, cause it looks like at around 16:20, tableclassmath seems to resort to trial and error.
Absolutely true. Well said.
He didn't solve it algebraically. He plugged in guesses after using algebra.
I did roughly the the same as you and ended up with (sqrt(3)^m + sqrt(2)^m) x (sqrt(3)^m - sqrt(2)^m) = 13 x 5. (Since 65 = 13 x 5.) Then I let sqrt(3)^m + sqrt(2)^m =13 and sqrt(3)^m - sqrt(2)^m = 5. Adding those two equations together gives 2xsqrt(3)^m = 18. Simplifying and squaring both sides yields 3^m = 81. Thus we know m is 4.
4
3^m = 65 + 2^m
Hence 3^m is at least 81 or 3^4 since m is an integer and since 65 is greater than 3^3 or 27
Hence m is at least 4
Try 4
3^4 -65 = 2^4
16 = 2^4
got 4 by brute force. great alternatives. thanks for the lesson.
Is the follwing method of fining answer acceptable to the Board? I resoned that 3^m has to be greater than 65 ...so first such power is 3^4 =81....I tried .... "81-2^4" is eactly 65!!!...... m =4.... IS THIS considered legal for high score?
It was not assumed that m is an integer. Even if that is the case, there is no reason for a=3^(m/2) and b=2^(m/2) to be integers. The only reasonable way to do this problem is to notice that for positive x, the function f(x)=3^x-2^x is increasing (hence injective) and therefore the solution will be unique - then we can get the solution by trial and error, or estimate the solution in case the answer is not nice (e.g., f(x)=64 also has a unique solution, but the answer is not an integer ). Then only interesting part of the problem would be to ask: where is the function f defined above is increasing? (Ans. for x>= -ln(ln(3)/ln(2))/ln(3/2) approximately -1.14).
With trial and error, I can get an answer, m=4. I of course still have to prove that this is the only solution
I can first say that, if m 0, and I can take the derivative wrt m of 3^m-2^m, giving me 3^m ln3 - 2^m ln2. Since ln3 >ln2 >0, 3^m ln3 > 2^m ln2, so the LHS is monotonically increasing wrt m. Therefore, m=4 is the only solution.
It's easy to demonstrate that 4 is an answer, but is it the only answer?
Answering that will require some algebra.
I was hoping to see an algebraic solution as well (if one exists) and found the trial and error solutions pretty disappointing. However, given that's what we have, couldn't we prove m=4 is a unique solution by simply noting that decreasing m always makes the right side smaller and increasing m always makes the right side larger? I'm sure there's a way to formalize that better, but hopefully, you get what I mean.
@@wrc1210 Sure there's a way to formalize that - it's called calculus.
@@jeffdege4786 For some reason YT suggested this video and I don't really want to spend time on the prolix solution presented in the video. Showing 3^x-2^x is monotonic increasing is a key step, but how to justify it without calculus? Perhaps start with the observation that 3^x is just 2^x but "squished" in the x-direction.
@@peterbrockway5990 3^x - 2^x = 2^x (1.5^x - 1). This is a product of two functions which are both monotonically increasing and which are both positive for x > 0. Which implies that the product itself also is monotonically increasing.
Is there an algebraic soloution available ? Same for 3^x - 2^x = 5
3^2 - 2^2 =5
The trouble with your “creative” method shown at ~15:00 is that by taking the square roots, you may end up with a and/or b which are not integers, even if m was. Then, you’ve no reason to assume that their sum and/or difference is an integer and therefore, the factorization of the right side is wrong as well. Indeed, if you try this with 3^m - 2^m = 19, you’ll end up with only one way to factor 19 as it is a prime: 19 x 1. This gives you a = 10 and b = 9. So now m should be log10/log(sqrt 3) as well as log9/log(sqrt 2)… Which isn’t even equal, let alone correct (as m = 3). I know you’re trying to show some less rigid, analytical methods, but as soon as you make any such assumption, you absolutely must note it, with a great exclamation mark and always check any result you obtain like that… I didn’t hear anything.
1. It's trivial 3^m > 65 . thus m>=4
2. (Bino. exp) 3^m-2^m = 65 = (2+1)^m-2^m = (2^m+m*2^(m-1)+.....+1) - 2^m = m*2^(m-1)+.....+1 > m*2^(m-1) , thus 2^7> 65 > m*2^(m-1), implies m
How to solve it when m is an odd number?
An excellent question! I was thinking the same thing! 🤔
Solution by insight
81-16=65
3^4-2^4=65
m=4
I just used guess and check. I knew it had to be a fairly small exponent because different bases would diverge very quickly, the difference is already pretty large at the exponent of 6.
So either trial and error the indices to start with, or trial and error a difference of squares?! If it's trial and error the first is easier. The difference of squares route shows working but it's not a proof because... it's trial and error. Or am i wrong?
I found by it by trial and error. First wanted to use small numbers to see if i could see a principle at play. Used 3, which is 27 - 8 = 19. So then I hoped 4 would be the right answer, and it was. But, I have not finished the video yet, so I hope there is a slicker way to solve this. Thanks.
Prepare to be disappointed.
@@wrc1210 Yes and No. Granted it was a very long way to get around the guess work...so in that sense, not very practical for me, anyway, but I did find it interesting...that it could be done. Did you watch to the end? Sometimes trial and error is the most efficient method in real life. That, in and of itself, perhaps is the most valuable lesson here.
@its-a-bountiful-life But he didn't "get around the guess work." He just moved it to a different spot after manipulating the equation a bunch and making it harder to do the guess work than it was in the original form.
I don't know. Sure, he demonstrated some algebra that might be useful in solving other equations, but it wasn't at all helpful in this one. Not a big deal, but kind of annoying. Wouldn't care as much if he was up front at the beginning and just said I'm not going to show you an algebraic solution to this, but here are some dos and don'ts about manipulating these types of equations and you'll see why this is such a difficult problem to solve without resorting to guess work.
This one hurt my brain. I need to recover now.
Could you please show how to solve this if the sum was 67 instead of 65? I’d like to see the solution for that one which didn’t depend upon guessology:
3^m - 2^m = 67
Solve for m
Well. getting b =4 goes to b= 2^ (m/2)= 4. No need work with a= 3^(m/2).
You are still first guessing b=l then guessing b= 4.
That's me: a little bit of basic maths and I can solve it! But I love learning the harder stuff. Just wish I could have followed it in school!!
What does M mean?
It's a variable
Here,
(m) is the variable form of power ( just like x or any other variable.) to the base 3 and 2
You can also set y = 0. And then solve the equation.. you'll get et 4.
Thanks for that. I originally started down all sorts of dead ends, then found the simple trial-and-error of method 1. However, I really enjoyed the last technique, even though that also involved some T&E.
You don't have to use T&E. For some reason John failed to complete his technique by using the factors of 65.
Is there no way to solve this without guessing? 😭😭
This approach is really confusing. The total lecture was 22:27 min. It took over 10 minutes in before anything of value.
Plot 3^m and 2^m. The difference as m increases is observed to increase. There is therefore only one real solution. Simple inspection reveals m=4.
There may be imaginary solutions.
at 17:49 we have to add a value to make the result a "better number". Why has that better number to be a perfect square?
a=2^(m/2). May be the original equation has also another solution, an odd value maybe. Then a is not a perfect square.
So it is not correct to use the fact that a has to be a perfect square, as you do in your video.
Maybe I did too much math ;-.) .. But I "see" .. 81 - 16 almost instantly (meaning to the power of 4 for both numbers) ^^^
But going the algebraic way .. at the point where you have (a+b)(a-b)=65 .. this calls for a product of 2 prime factors 13 and 5 ... which yields 9 and 4 pretty quickly which gives m=4 in return
I tried doing it a different way using logs and I got 2 different answers 3.700 and 4.41. What am I doing wrong?
Divide the terms on the LHS by 2^m
If you do 2^m([3/2]^m - 1)= 13 x 65. (Just to check, 2^m x (3/2)^m = 3^m)
2^m = 13; mlog2 = log13; m= log13/log2 and I get 3.700
(3/2)^m -1 = 5; (3/2)^m = 5 +1; (3/2)^m = 6; mlog1.5 = 6; m = log6/log1.5; m = 4.41.
Please let me know where I have gone wrong😂. My maths has gone
no need for any formula , we deal with small figures: 3 and 2 at forth exponent
3^m - 2^m = 65
I remember that 81 - 16 = 65, so I can rewrite them as powers of 2 and 3:
3^4 - 2^4 = 65
Thus, m = 4 .
There's no need to step through integers 0, 1, 2, 3, ..., for 3^m - 2^m - 65 = 0 unless there's no integer solution. In that case, we can find where the potential m value is between two successive integers that straddle 0, if at all. If no crossovers are found, then there's no solution.
I knew the factors of 65 as 5 & 13, and if I used the difference of squares, the factors are the difference and sum of 3^(m/2) and 2^(m/2). The average of the factors is 3^(m/2) and 9, so m = 4. As a check, half the difference is 2^(m/2) and 4, which also has m = 4.
Still feels like a substitution cheat. How is this more eloquent then simply iterating from the get go? Was hoping for an analytical solution.
If you get to (x+y)(x-y)=65 then factor to (x+y)(x-y)=13x5 then break it out x+y=13 and x-y=5, you can solve for each and find that x=9 and y=4, and you're there.
Trial and error. Or graph on your calculator y=3^x-65 and y=2^x . Where the curves intersect is the answer on the x axis.
That's not solving, that's guessing
He’s the teacher stop hating on him
M=4
@@John-PaulMutebiIt's a terrible math problem.
I wouldn't say it's "guessing", more like brute force try and see. Since you don't try random numbers
Okay, just trial and error. But how do I solve this for non-integers?
If m is negative then 3^m and 2^m are in [0, 1] and their difference cannot be equal to 65, so if the given equation has solutions they are positive.
f: R+ ---> R+, x ---> 3^x - 2^x strictly increases on R+ (positive derivative), so if the given equation has a positive solution then it is unique.
As x = 4 is evident solution, it is the only positive one and finally the only real solution.
m=4. 5 seconds to check on my arithmetic
I once had to solve a fluid mechanics problem in an exam, where one first had to derive the formula for the flow rate through a sluice gate and then solve the formula to obtain the flow rate. Once derived, the equation for the flow rate could not be solved by algebra and most students gave up and moved on to the next question, thinking that the formula they had derived must be wrong. After the exam, the professor explained that there was no need to use algebra to obtain the answer and that a numerical solution would have been acceptable. In this case, I would set the equation to zero, (3^m-2^m)-65=0. Then I would find a trial value for m that gives a positive answer to the equation and a trial value for m that gives me a negative answer. I would then find the root of the equation for zero by using a numerical algorithm such as the bisection method for example. In this case, it was not necessary for me to do this because I found the root of the equation while I was searching for two suitable trial values of m. If the answer was not a whole number, it would have been different.
I'm not happy with trial and error approach used here. Did some searching and this seems to be solvable by using the Lambert W Function. ChatGPT almost shows how to do it. Will the future math classes be taught by AI?
Good golly - so many words! Like those web pages that draw you on and on to keep you scrolling through all their adverts, to the trivial end. Never again. Unless I need something to put me to sleep.
It's easy enough to take the second term to the other side and work with increasing 65 by 2^m.
A BIG thanks to tabletclass math for what you do.
I personally call you Jay-z.(John Zimmerman).
I appreciate you.God bless you.
1:56 " . . . you're taking an *algebaric* perspective . . . "
Surely you mean ---> an *algebraic* . . . whatever it is
That went down well!
I tried the log method and blew it. I'm changing majors. Maybe to art history or gender studies.
It means you do not understand math. You can see right away that LOG would not work.
@@danv2888You're correct. I don't. That's why I just switched to gender studies. My first assignment is defining what a woman is. I can solve that one.
As the narrator stated in the video, the equation does NOT lend itself to solution by the rules of logarithms. It *looks* like it does, but it does not. You can't simplify log (3^m - 2^m)
The quick way is just to put it on a spread sheet and try some values of m. I guessed 4 and hit it first try.
you are really good
Not realy because m= natural number 1,2,3,4 and so on and you coul solve by trial and error method.
Where is it stated that m is a natural number?
3^m - 2^m=65=81-16
3^ 4- 2^4 m=4
There was no need for checking for value of m for 2 till the value of 3 does not exceed 65.
I would solve it using modulos. First mod 3 shows m even: m = 2a. 9^a - 4^a = 65 mod 8 gives a even: a = 2b. Then: 81^b - 16^b you can always factorize: 81^b - 16^b = (81-16) (81^(b-1) + 81^(b-2)*16 + ... + 16^(b-1))
This sum is always a positiv integer. Therefor 81^b - 16^b >= 81-16 = 65. And equality only if the above sum is 1, which is the case only for b = 1, and thus n = 4.
You still did some guesswork to obtain the algebraic solution.
Trial and error.
First guess must be 4 (since 3^3 is 27 - not big enough)
3^4=81. 2^4=16. 81-16 = 65. DONE
Took me 20 seconds because I still count on my fingers.
y= 4. why do "most won't figure out how to solve"?
For this problem, graphing is the only real solution that doesn’t require trial and error.
It's easy to "cheat" and get the right answer in a few seconds by trying, "1,2,3,4" in your head, but if the answer was not a simple integer, the next level of "cheating" would be to observe that 3^m grows faster than 2^m, so the function will grow monotomically with "m", and you can write a computer script to do a binary search, That may still be "cheating", but practicing with such scripts can be a valuable skill for its own sake, extendable to other kinds of problems that don't lend themselves to other methods.
But still spend the 22 minutes to learn the algebraic methods, as well.
The only answer I can get by conventional means are two values of m: 4(log3{base 3} and 4(log2{base2}
Update: I find that both values =4 numerically
Great, learned a lot
The second one I tried in my head: 4. How hard was that.
When author got a^2-b^2=65, after this he overcomplicated. I will recommend (a-b)*(a+b)=5*13; Therefore, a-b=5 and a+b=13, 2*a=18: a=9: b=4; When a-b = 13 and a+b=5: 2*a=18; a=9; b=-4; If (a-b)*(a+b)=1*65; then a-b=1, a+b=65 or a-b=65 and a+b=1. When we solve a=33, b=32 and a=33; b=-32.
Great. That is the way to do it. The guy does not know how to solve this problem. He is doing a disservice to all who want to learn.
Not sure about the algebra without some head scratching, but got the answer in about 10 secs with trial and error.
I started with M = 3 and it didn’t work.
(3x3x3)= 27 - (2x2x2) = 8 so wrong answer of 19
Then M=4 it worked
3x3x3x3 = 81
2x2x2x2 = 16
82-16= 65
I did this one thanks to a lucky guess to be quite honest.
m had to be 'more than 3' because 3^m had to be bigger than 65
and 4 came next ..
3^4 - 2^4 = 81 - 16 = 65
Mental arithmetic in 30 seconds. 3x3x3x3=81. 81-65=16 = 2x2x2x2. So m = 4.
C'mon, I did this in my heat in about 10 seconds.
He is describing an algebraic principle so can solve more complex problems
It helps a lot just to know the solution is integer.