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.

@jim2376

2 ай бұрын

👍 Exactamundo!

@stevespencer4445

2 ай бұрын

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...

@royschering1140

2 ай бұрын

@@stevespencer4445 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

@margaretcorfield9891

Ай бұрын

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

@pauladams9893

3 күн бұрын

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.

@kade82

2 ай бұрын

I solved it in my head using trial and error in seconds. How to solve it "correctly," I'm not sure.

@AlainPaulikevitch

2 ай бұрын

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.

@wrc1210

2 ай бұрын

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.

@zemethius

2 ай бұрын

Solving this algebraically would need logarithms, at which my brain just goes "no" and walks away. Trial and error worked for this case.

@LyneisFilm

2 ай бұрын

@@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.

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.

@stevespencer4445

2 ай бұрын

"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...?

@Pal4you1946

Ай бұрын

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

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

That's not solving, that's guessing

@user-or6bk2wt1o

2 ай бұрын

He’s the teacher stop hating on him

@edithsmith1524

2 ай бұрын

M=4

@eccentricaste3232

2 ай бұрын

​@@user-or6bk2wt1oIt's a terrible math problem.

@OrestesKyriakosPoulakis

28 күн бұрын

I wouldn't say it's "guessing", more like brute force try and see. Since you don't try random numbers

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.

what we used to call the guessology technique

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

@88kgs

2 ай бұрын

Yes

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

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.

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

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 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.

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 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.

@warblerab2955

2 ай бұрын

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.

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

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.

@vespa2860

2 ай бұрын

You don't have to use T&E. For some reason John failed to complete his technique by using the factors of 65.

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.

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

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

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

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 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.

@nickcellino1503

2 ай бұрын

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.

@KipIngram

2 ай бұрын

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 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).

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.

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.

In both examples you simply plugged in numbers that worked.

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, learned a lot

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

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.

@letsfinishit5484

2 ай бұрын

I think he should leave it in the title.

@Pax.Alotin

2 ай бұрын

At least he is avoiding clickbait titles like ---- 'This destroys most Mathematicians' -- or --- 'What Scientists tried to hide' ---- 😎

@untouchblz

2 ай бұрын

Are you getting them all wrong?

@stompthedragon4010

2 ай бұрын

That actually makes me glad to be one of the collective instaed of feeling like a dunce in the minority.

@stevespencer4445

2 ай бұрын

Or maths even...

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

2 ай бұрын

Prepare to be disappointed.

@its-a-bountiful-life

2 ай бұрын

@@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.

@wrc1210

2 ай бұрын

@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.

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.

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

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

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

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.

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

@eccentricaste3232

2 ай бұрын

3^2 - 2^2 =5

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.

@oahuhawaii2141

2 ай бұрын

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.

@wrc1210

2 ай бұрын

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

2 ай бұрын

@@wrc1210 Sure there's a way to formalize that - it's called calculus.

@peterbrockway5990

22 күн бұрын

@@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.

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.

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

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

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

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

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.

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

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

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.

A little basic guess ‘n’ check solves this

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

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

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.

I FIGURED IT OUT IN MY HEAD. 81-16

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

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

@wrc1210

2 ай бұрын

Where is it stated that m is a natural number?

What does M mean?

@louf7178

2 ай бұрын

It's a variable

@88kgs

2 ай бұрын

Here, (m) is the variable form of power ( just like x or any other variable.) to the base 3 and 2

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.

@wrc1210

2 ай бұрын

Sometimes, common sense can mislead us. Try: 3^m - 2^m = 66 I think you will find m is definitely not an integer.

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

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

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

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

@danv2888

2 ай бұрын

It means you do not understand math. You can see right away that LOG would not work.

@terry_willis

2 ай бұрын

@@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.

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

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

@warblerab2955

2 ай бұрын

but it is not just algebra. It is algebra that leads to guessing and checking as far as I can tell.

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.

@stevespencer4445

2 ай бұрын

Why obviously an integer just because 65 is? Subtracting a non-integer from a non-integer also gives an integer....

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

@peterbrockway5990

22 күн бұрын

What's the log of the LHS?

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

Excel will suffice for a calculator

Hint: 81 - 16 = 65

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

I imagine your classes being 12 hours long. Stop guessing how people feel and get to the point.

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.

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

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.

I like that one

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

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

4

m+4

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?

Allow myself to introduce myself

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

No real mathematics were involved here

Solve it in two minutes

Heuristic

Amazing how many things most people get wrong

2_3/45❤

I hate to say this since you put a lot of work into this video, but I don’t see the “guess and check” method as a valid “solution” to the problem. Sure, the correct answer may be 4, but if you had to figure it out without guessing, what kind of answer would you come up with. Obviously, we know the answer must be at least 4 (though it could theoretically be 3 1/2), because 3^3 is 27, which is less than 27, and 3^4 is 81. That is true, but there needs to be a legitimate way to prove this algebraically. If a student gave this kind of solution on a math test, they would get a maximum of 2 out of 10 points, even though they arrive at the right answer, because the solution isn’t a “legitimate solution”.

@untouchblz

2 ай бұрын

Guess and check is a valid method.

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

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.

By eye because I got the 65 and 81...😅

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

The power is 4

M os the same power. Thos os algebraic!