3^m - 2^m = 65 MOST won’t FIGURE OUT how to solve!
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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.
I would suggest another way to find out the solution.
After arriving to (a+b)(a-b)=65, we noticed that 65 can only be obtained multiplying 65 by 1 or 13 by 5.
If we assume a+b=13 and a-b=5, after adding both equations we have 2 x a = 18, then a=9.
If 3 powered to m/2 iquals to 9 (or 3 powered to 2) then m/2= 2 and m=4.
We can solve it in another way from the concept of a² - b² = (a+b)(a-b). For example, we we have already written(3^m/2)² - (2^m/2)² = 65. Since this is in the form of a² - b² = (a+b)(a-b), we can write this as (3^m/2 + (2^m/2)(3^m/2 - 2^m/2) = 65. Considering m as a positive integer, the LHS is the product of two terms. Also (3^m/2 + (2^m/2) > (3^m/2 - 2^m/2). Again 65 is the product of (i) 13 and 5 and (ii) 65 and 1. There is no third pair other than these two. Therefore, we two cases, (a) (3^m/2 + (2^m/2) = 65 and 3^m/2 - 2^m/2) = 1 and (b) (3^m/2 + (2^m/2) = 13 and (3^m/2 - 2^m/2) = 5. Solving the first set (a), that, adding them, we get 2. (3^m/2) = 65+1=66, that is 3^m/2 = 33. We cannot get a solution for this, since no power of 3 is 33. Solving the second set (b) in the same way, we get 2. (3^m/2) = 13+5=18, which gives 3^m/2 = 9 =3². From this we get m/2 =2, meaning m =4.
"Considering m as a positive integer"? Why? Who says you can just consider something to be something. What if it's not 65, but instead a value that means m is not an integer...?
In Europe, in the 60's from 6th grade to 12 th grade each subject was graded with numbers. Not letters. Graded out of 20. And you needed to have 10 to pass. All grades of each subject were added every month and divided by the numbers of tests. But each subject had a coefficient, meaning some were worth more than others. Like in college. Some courses get 3 credits some get 4. So if you got an 8/20 in english for instance and a 16/ 20 in history you would add the extra 6 to English and you passed. Well...
For 7 years I had 1/20 in math sometimes 1.5/20. Hahaha. And I graduated with a 12/20
Then I went to college and only took one math course Algebra 1.
Thank you. But I still don't get in. Lol 14:56
In particular, you have reduced the problem to solving two linear equations in two unknowns.
I was keeping up until about @16:20, when it looked to me like he just decided to pick random numbers for b.
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
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.
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.
You can also set y = 0. Amd them solve the equation..
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.
In both examples you simply plugged in numbers that worked.
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
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.
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. Multipy 3m by the other number then subtract 65 and use pi. The other M is not used. I go to a school in Oregon that teaches if you try you will be correct even if wrong. So I am correct here.
It's easy enough to take the second term to the other side and work with increasing 65 by 2^m.
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
There was no need for checking for value of m for 2 till the value of 3 does not exceed 65.
I did this one thanks to a lucky guess to be quite honest.
m had to be 'more than 3' because 3^m had to be bigger than 65
and 4 came next ..
3^4 - 2^4 = 81 - 16 = 65
3^m - 2^m=65=81-16
3^ 4- 2^4 m=4
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.
After fifteen minutes of going over all the wrong ways to do the problem you used difference of squares and substitution. This works on the assumption that the answer is a whole number (integer). Was that restriction included in the beginning?
Is there an algebraic soloution available ? Same for 3^x - 2^x = 5
3^2 - 2^2 =5
The second one I tried in my head: 4. How hard was that.
3hoch m muss größer sein als 65
Also probieren wir m =4
3hoch 4 - 2hoch 4 = 81 - 16 = 65
Ready
What does M mean?
It's a variable
Here,
(m) is the variable form of power ( just like x or any other variable.) to the base 3 and 2
you are really good
After solving we get m=4
27min to say "by trials and errors", I can't believe!!!
3 and 2 at the fourth power
All good and fine if m is a whole number. How about 3^m-2^m=64? Here only algebra can be used 😅😅
sorry - approach to solution isnt mathematics .. if we assume b=1 and then working it for a perfect square after addition to 65 then why not just assume m=1 then m=2 then m=3 and so on .. why do all of these additional steps ? for what reward . what if its not 3 and 2 - its 10345343 and 327773563 or root of 3 and root of 2 then how do you solve this .. that is math
If only this video had shown us "m=4" is the only solution, real or complex.
Hint: 81 - 16 = 65
Its confusimg bc you are all over the place! Yeach that of you see this kind pf equation, know ot as a Y = problem and set Y = to o..them dove the eqation!
m= 4
M=65
1 ×1=1 1+1=3 now you know the basics everthang else is eezee
m = pie. (Cherry pie)
Guestimates ...to "move closer" to a possible solution "area" only works in the simplest of scenarios.
That's why basic log mapping would have at least given the target area in general - and then mapping in an app would refine it to exactitude
Really did not enjoy this video , because it did not express the approach at all well.
Solve it in two minutes
4
Among the numerus people who cannot solve this equation, count the guy who does the video. One must solve it without writing anything...m obviously bigger than 3. 4 works. End of it.
waaaay too much gab!
m=4. That took all of 10 seconds to figure out!
No easy way to compute this.
Should take less than 10 seconds to realize the answer in your head. Is the 1st part 27 or 81? Ok, 81, obviously. Ok, it's 4. Christ.
You just viloted pemdas
Dabba question
Mighty long explanation to not really have the answer. You're not really a teacher are you.
With your long-winding monologue and scribbling and doodling you stretch your time on youtube, and also stretch/spaghetify the poor not-so-crazy-about-math kinds, traumatizing them. Poor kids😢😮
Too easy.m=4.
Even Trump can do this
Answer m=4😂
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.
m=4
M=65
4
To me, the easier approach was to go above the 65 with the 3m power because we will subtract 2m power back to the answer of 65. 3 to the 3rd power is only 27 and not above 65, so 3 to the 4th power is 81. 81-65=16=2m power or m=4. So it all plugs in. I don't know if I could remember how to do it in the video without seeing the entire thing on paper to follow.
That was my method also. It did not take long and is more straightforward than the algebraic approach.
what we used to call the guessology technique
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.
I'm trying to solve this with logs, but it isn't working. It should work, yeah? Please help!!
This lesson jumped to the highly extraordinary, but is a good example of why the sum of exponents can't simply distribute the exponent.
I wish you would make another utube on the different ways of solving this equations thanks
Yes
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)
m=4. 5 seconds to check on my arithmetic
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.
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.
what about 65 = 13*5 or 65*1 -> 4 cases for (a - b)(a + b)
Is there no way to solve this without guessing? 😭😭
This approach is really confusing. The total lecture was 22:27 min. It took over 10 minutes in before anything of value.
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?
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.
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
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. :)
C'mon, I did this in my heat in about 10 seconds.
He is describing an algebraic principle so can solve more complex problems
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.
Not realy because m= natural number 1,2,3,4 and so on and you coul solve by trial and error method.
Where is it stated that m is a natural number?
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.
Why using logarithm does not work with this equation ?. For example Log3m-Log2m = Log65.
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
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
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.
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.
This video is a mess. The worst math channel on TH-cam. I'm checking out, 👋
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.
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!
You still did some guesswork to obtain the algebraic solution.
Mathematicians doing this kind of elaborate, overcomplicated method of solving an equation that is much simpler and faster to solve by TRIAL AND ERROR, is why I've hated math my entire life.
Getting the correct answer never counted if we didn't do it the way they wanted us to do it, which never made sense to me.
I watched the whole video thinking I might learn something, but as far as I'm concerned, the "right" solution was way over my head, and I simply don't see the point.
You didn't solve this problem using algebra. You used numerical analysis. You guessed at the answer like your confused students do. Yes you can also use a computer to do the guessing which can guess much faster and also get some precision too. Yes, I also know this problem can be solved using higher math that is way beyond your target audience of 9th grade math skills. Never, ever tell your students to guess at the answer as a way of doing math. Guessing is just a way to answer problems on tests. You shouldn't be guessing in math class. Math class is to learn math, not guess at the answer. So now what are you going to do when the problem 3^m - 2^m = 67 and not 65. What is the answer? Yes we know the answer is between 4 and 5. I want the answer to 10 decimal places.
Great lesson .but you are going too fast in the critical areas. Leave out the confused faces sketches with its commentary, please.
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+...
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”.
Guess and check is a valid method.
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.
It's unbelievable: 22 minutes of non-math. Just pure tryout by logorhea. - Sorry!
Owe, i didnt see m=4. Lol. Im pretty tired..if m = 4, it's the same as pemdas. Exponents first..or ate you asking what is m?
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
got 4 by brute force. great alternatives. thanks for the lesson.
You should be able to do it in your head quickly.
65 is the result, so start with 3= m...nope. 4= m...yes.
Love your work, bur boy you overcomplicated this, Might help to back off on the rhetoric,
no need for any formula , we deal with small figures: 3 and 2 at forth exponent