3^m - 2^m = 65 MOST won’t FIGURE OUT how to solve! 

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.

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 was keeping up until about @16:20, when it looked to me like he just decided to pick random numbers for b.

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.

This lesson jumped to the highly extraordinary, but is a good example of why the sum of exponents can't simply distribute the exponent.

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

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.

That's not solving, that's guessing

what we used to call the guessology technique

I'm trying to solve this with logs, but it isn't working. It should work, yeah? Please help!!

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.

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 wish you would make another utube on the different ways of solving this equations thanks

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.

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.

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.

This one hurt my brain. I need to recover now.

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.

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.

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.

What would be the algebraic solution for 3^m - 2^m = 2 for example?

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

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.

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.

Great, learned a lot

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.

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 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.

Mental arithmetic in 30 seconds. 3x3x3x3=81. 81-65=16 = 2x2x2x2. So m = 4.

One shouldn't discount taking the easy route with problems -- when the possibility is there. 3^m grows MUCH faster than 2^m. m is probably small. 3^3 =27. Nope, 3^4 = 81 (3x27). Hmm. 81-16? Yup. It's 65' m=4. Time to calculate in my head: maybe 6-7 seconds. Works for for exams where time is a factor.

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.

got 4 by brute force. great alternatives. thanks for the lesson.

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.

To find m at the end one can use log, too.

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).

One word of warning while there is nothing wrong with this technique: you need to prove that more solutions does not exist, and if they do solution like this is incomplete.

In both examples you simply plugged in numbers that worked.

You can also set y = 0. Amd them solve the equation..

A BIG thanks to tabletclass math for what you do. I personally call you Jay-z.(John Zimmerman). I appreciate you.God bless you.

Let f(m) = 3^m - 2^m where m is real. Clearly there is no solution where m < 1. Now when m >= 1, then f(m) is always increasing. Now as f(1) = 1 < 65, then this means f(m) = 65 has only one real solution. We observe that f(4) = 65 and we have found a solution and it is the only solution and we are done.

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 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

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.

Easy assume m=2k then use a^2-b^2. Factor 65 = 13×5

Feel like I was calf roping....I yelled Time when I had it 🤠

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.

Is there an algebraic soloution available ? Same for 3^x - 2^x = 5

It helps a lot just to know the solution is integer.

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.

You can also set y = 0. And then solve the equation.. you'll get et 4.

m =. 4 m. m 3. - 2. =. 65 3x3x3x3. - 2x2x2X2 81. - 16. =. 65 For me...Less than 5 secibds

Is there no way to solve this without guessing? 😭😭

m=4 solved in my head in 13 seconds, and is only solution because of the laws of exponents

It's easy to demonstrate that 4 is an answer, but is it the only answer? Answering that will require some algebra.

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.

It's easy enough to take the second term to the other side and work with increasing 65 by 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.

27min to say "by trials and errors", I can't believe!!!

A little basic guess ‘n’ check solves this

I tried the log method and blew it. I'm changing majors. Maybe to art history or gender studies.

Harder algerbra but good learning experience. Remember, you don't get stronger unless you get outside of your comfort zone.

It's obviously m=4. It has to be an integer, since the difference is an integer. It takes no more than a few seconds to see that 4 gets you there.

Got it in five seconds! Why does it need such a long video?!

I solved it using trial and error; by the time I got to M=4 I got the solution. I figured there had to be an easier way.

Common sense says that an integer to a power minus and integer to the same power equals an integer - chances are the common exponent will be an integer. Since the integer must be greater than 3 to render an number greater than 65, the first possibility is 3^4, 81. That would make 2^4, or 16. 81 - 16 = 65, which is the desired result, so m = 4. You can do this in your head.

I FIGURED IT OUT IN MY HEAD. 81-16

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

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 like that one

All good and fine if m is a whole number. How about 3^m-2^m=64? Here only algebra can be used 😅😅

assumes m is an integer without telling that to the solver means guessing or graphing is just a shot in the dark. not a math solution

1:56 " . . . you're taking an *algebaric* perspective . . . " Surely you mean ---> an *algebraic* . . . whatever it is

Still feels like a substitution cheat. How is this more eloquent then simply iterating from the get go? Was hoping for an analytical solution.

And you wonder why us poor unwashed have difficulty understanding math. Everything is a conundrum. It's kind of like, "you do it this way, unless it is a Tuesday in April of a leap year which falls on the week before Easter." In other words, it' all gobbledygook that appears to be coming out of nowhere. Where is the system of thought which leads us us to a step by step process that leads us to the proper answer?

If we take logarithms of both sides and solved, we get an entirely different answer why?

Not realy because m= natural number 1,2,3,4 and so on and you coul solve by trial and error method.

Seriously? Yes it can be done by trial and error with easy values. But i thought he was "showing working" like exams like. But when he started letting a value = 1, i was like, huh? How is that even a technique? I was lost at that point...and rewinding through all the waffle was too hard

I think it's time to get rid of the "most will get it wrong" tag line. It's a bit redundant. Most people can't do math.

4

Excel will suffice for a calculator

You should be able to do it in your head quickly. 65 is the result, so start with 3= m...nope. 4= m...yes.

2_3/45❤

m=4 done in my head, so I wasn't sure it was exactly correct

Hint: 81 - 16 = 65

m+4

Heuristic

U can plug in numbers and solve it in 2 minutes.

I’d use numerical analysis techniques and let the computer solve it.

Good thing my life doesn’t depend on a correct solution. 😜

wow! that was the most difficult problem I've ever seen the you tube math man take us through.

You really lost me on this one. Thanks for the opportunity.

I usually look at your videos daily to refresh the math I learned 60-70 years ago. This video was terrible since the obvious guess as the beginning was made a complicated guess at the end.

Solve it in two minutes

What does M mean?

m=4. That took all of 10 seconds to figure out!

Amazing how many things most people get wrong

The power is 4

Keep up the good work john. Its getting people to think in the right direction. I did thermodynamics and chemistry. you know how we figured out how much chemicals to add to reach a desired concentration? swag. scientific wild a"" guess. we did basic rounding math and then added a best guess. perfection was not a goal when you have an acceptable range of 10-50 ppm. Thanks for your hard work putting these videos together.

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.

You are assuming, m is even and a and b will be real numbers.