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
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.
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.
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.
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
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.
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.
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.
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. :)
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 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.
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.
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.
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.
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
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
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.
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.
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!
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 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.
I'll give credit where credit is due. I didn't think to characterize the problem as a difference of two squares. Thank you for doing the heavy lifting for me. Thanks to your insight we have: (3^[m/2] + 2^[m/2])×(3^[m/2] - 2^[m/2]) = 65. The prime factorization of 65 = 13×5. Therefore (3^[m/2] + 2^[m/2]) = 13 and (3^[m/2] - 2^[m/2]) = 5. Adding these two equations together gives: 2·3^[m/2] = 18. Dividing by 2 on both sides, 3^[m/2] = 9. Finally m/2 = log(9)÷log(3) = 2. Therefore, m = 4 ◼
The solution assumes that m is an even integer, otherwise it wouldn't make sense to use a prime factorization. So in order to show that is the only solutions, you have to do more work.
@@bjornfeuerbacher5514 Actually it doesn't assume anything about the parity of m. Also the prime factorization of any given number is unique. The prime factorization of 65 = 5 × 13, has two factors, 5 and 13. Likewise, 3^m - 2^m written as a difference of two squares has two factors; hence, one of them is equal to 5, and the other is equal to 13.
@@johnnolen8338 My point is that if m is not an even number, then 2^(m/2) and 3^(m/2) are not integers, they are irrationals. So then you have the condition that the product of two irrational numbers is 65. The prime factorization of 65 hence would not help in that case.
@@bjornfeuerbacher5514 You make a good point. However, number theory deals exclusively with properties of integers. This is a number theory problem. (Which can also be proved in itself, but the proof is beyond the scope of a typical Algebra II course.) Inasmuch as this is a number theory problem, there is the unstated a priori acknowledgment that the solution m will be an integer. Notice that I didn't say it was an assumption, because it's not assumed. It's a fact. However, all of that speculation; i.e. what if m is not an integer?" is put to rest by the observation that we found the solution, m = 4 in this case, and this is the only solution! ◼
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.
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.
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.
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.
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.
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).
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
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.
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.
4. To write down the answer you need to already know that 3^4 is 81 and 2^4 is 16. If you don't know that the problem is too hard for you. It's not meant to be rocket science.
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.
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.
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.
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.
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.
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 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.
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.
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.
One word of warning while there is nothing wrong with this technique: you need to prove that more solutions do not exist, and if they do solution like this is incomplete.
Если есть очевидное решение, то проще всего доказать, что оно уникально. Решение m равно четырем. Производная от левой стороны больше. Поэтому решение уникальное.
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.
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.
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
Okay so power of 3 are 9, 27, 81 and powers of 2 are 4, 8, 16. Since 81-16=65, at least one solution is m=4. But this is not a rigorous solution. Now to watch!
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
@@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)
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
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.
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.
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?
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.
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.
This lesson jumped to the highly extraordinary, but is a good example of why the sum of exponents can't simply distribute the exponent.
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.
I was keeping up until about @16:20, when it looked to me like he just decided to pick random numbers for b.
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.
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. :)
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 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.
what we used to call the guessology technique
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.
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.
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.
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
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
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.
I wish you would make another utube on the different ways of solving this equations thanks
Yes
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.
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!
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 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.
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
In both examples you simply plugged in numbers that worked.
What would be the algebraic solution for 3^m - 2^m = 2 for example?
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.
I'll give credit where credit is due. I didn't think to characterize the problem as a difference of two squares. Thank you for doing the heavy lifting for me.
Thanks to your insight we have: (3^[m/2] + 2^[m/2])×(3^[m/2] - 2^[m/2]) = 65.
The prime factorization of 65 = 13×5. Therefore (3^[m/2] + 2^[m/2]) = 13 and (3^[m/2] - 2^[m/2]) = 5. Adding these two equations together gives: 2·3^[m/2] = 18. Dividing by 2 on both sides, 3^[m/2] = 9.
Finally m/2 = log(9)÷log(3) = 2. Therefore, m = 4 ◼
John, you have the best solution. Your solution earns an A+...
The solution assumes that m is an even integer, otherwise it wouldn't make sense to use a prime factorization. So in order to show that is the only solutions, you have to do more work.
@@bjornfeuerbacher5514 Actually it doesn't assume anything about the parity of m. Also the prime factorization of any given number is unique. The prime factorization of 65 = 5 × 13, has two factors, 5 and 13. Likewise, 3^m - 2^m written as a difference of two squares has two factors; hence, one of them is equal to 5, and the other is equal to 13.
@@johnnolen8338 My point is that if m is not an even number, then 2^(m/2) and 3^(m/2) are not integers, they are irrationals. So then you have the condition that the product of two irrational numbers is 65. The prime factorization of 65 hence would not help in that case.
@@bjornfeuerbacher5514 You make a good point. However, number theory deals exclusively with properties of integers. This is a number theory problem. (Which can also be proved in itself, but the proof is beyond the scope of a typical Algebra II course.) Inasmuch as this is a number theory problem, there is the unstated a priori acknowledgment that the solution m will be an integer. Notice that I didn't say it was an assumption, because it's not assumed. It's a fact.
However, all of that speculation; i.e. what if m is not an integer?" is put to rest by the observation that we found the solution, m = 4 in this case, and this is the only solution! ◼
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.
Take log on the both sides. M can be determined
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
Solution by insight
81-16=65
3^4-2^4=65
m=4
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.
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 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.
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.
got 4 by brute force. great alternatives. thanks for the lesson.
what about 65 = 13*5 or 65*1 -> 4 cases for (a - b)(a + b)
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.
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.
A BIG thanks to tabletclass math for what you do.
I personally call you Jay-z.(John Zimmerman).
I appreciate you.God bless you.
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).
m=4. 5 seconds to check on my arithmetic
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
This one hurt my brain. I need to recover now.
For this problem, graphing is the only real solution that doesn’t require trial and error.
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.
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.
You can also set y = 0. And then solve the equation.. you'll get et 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.
4. To write down the answer you need to already know that 3^4 is 81 and 2^4 is 16. If you don't know that the problem is too hard for you. It's not meant to be rocket science.
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.
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.
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.
M = 4th power =3m-2m=65 = (3×3×3×3)-(2×2×2×2)= (81)-(16)=65 That was easy.
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.
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.
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.
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 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.
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.
It's easy enough to take the second term to the other side and work with increasing 65 by 2^m.
3^m - 2^m=65=81-16
3^ 4- 2^4 m=4
3^4-2^4=65.....m=4
=>(3^m/2+2^m/2)(3^m/2 -2^m/2)=13(5)
=>3^m/2=9=3^2.... =>m=4
also 2^m/2=4.....=>m=4
Confirmed
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.
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
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.
I think he should leave it in the title.
Are you getting them all wrong?
That actually makes me glad to be one of the collective instaed of feeling like a dunce in the minority.
Or maths even...
Relax Neon his name tells you where he spends most of his time using sandpaper instead of toilet paper
One word of warning while there is nothing wrong with this technique: you need to prove that more solutions do not exist, and if they do solution like this is incomplete.
A BIGGG YESSSS
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.
m =. 4
m. m
3. - 2. =. 65
3x3x3x3. - 2x2x2X2
81. - 16. =. 65
For me...Less than 5 secibds
Harder algerbra but good learning experience. Remember, you don't get stronger unless you get outside of your comfort zone.
but it is not just algebra. It is algebra that leads to guessing and checking as far as I can tell.
no need for any formula , we deal with small figures: 3 and 2 at forth exponent
27min to say "by trials and errors", I can't believe!!!
Is there no way to solve this without guessing? 😭😭
Mental arithmetic in 30 seconds. 3x3x3x3=81. 81-65=16 = 2x2x2x2. So m = 4.
Can you please make a video to solve 3^m - 2^m = 66?
Если есть очевидное решение, то проще всего доказать, что оно уникально. Решение m равно четырем. Производная от левой стороны больше. Поэтому решение уникальное.
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.
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.
Sometimes, common sense can mislead us. Try:
3^m - 2^m = 66
I think you will find m is definitely not an integer.
I have a friend who says math is dark magic. With all the steps involved in this video, I would agree with her
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.
3x3x3x3=81 et 2x2x2x2=16, 81-16=65. m=4
Show your work. You just guessed the answer was 4.
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
Still feels like a substitution cheat. How is this more eloquent then simply iterating from the get go? Was hoping for an analytical solution.
Easy assume m=2k then use a^2-b^2. Factor 65 = 13×5
you are really good
Okay so power of 3 are 9, 27, 81 and powers of 2 are 4, 8, 16. Since 81-16=65, at least one solution is m=4. But this is not a rigorous solution. Now to watch!
That went down well!
Great, learned a lot
This approach is really confusing. The total lecture was 22:27 min. It took over 10 minutes in before anything of value.
65=13x5
2xa^2=18 a=3^m/2
a^2=9. m=4
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
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.
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)
Is there an algebraic soloution available ? Same for 3^x - 2^x = 5
3^2 - 2^2 =5
There was no need for checking for value of m for 2 till the value of 3 does not exceed 65.