Math Olympiad 3^m–2^m=65 | Math Olympiad Problems | Algebra

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In solving this math Olympiad problem, 3^m–2^m=65, Jakes uses a very unique approach to handle this exponential math challenge with easy.
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The comment section is polluted by critics but i learned some tricks from this video.. EXCELLENT

tiehoteele

I worked it out a bit different. My solution was simply determine what of 3 exponent would get me a number greater than 65 that would be an odd number (3^4). I then subtracted that 65 from that number (81) and I got 16 which is 2^4.

In other words you can rewrite the equation in this instance as

3^m - 2^m=65
3^m - 65 = 2^m
The first exponent of 3 which results in a number greater than 65 is 4
so 3^4 = 81
81-65 = 2^m
16 = 2^m
16 can be written as 2^4
m=4

danieldavies

I was impressed by the analytic demonstration used to figure out that m = 4. Sometimes procedures can be more interesting to follow along than just knowing the result.

alfredomulleretxeberria

I immediately saw that 65 can be decomposed as 81-16, and conveniently 81 is 3^4 and 16 is 2^4, so matching coefficients suggests m is 4.

mitahaubica

There is a problem here, when you write (x+y)(x-y)=5x13 you can not deduce that x+y=13 because you do not know if x+y is a natural number. If m is an odd number, then x=3^(m/2) and y=2^(m/2) are not natural numbers. So you have prove that if m is an even number, then m=4. it is very easy to check that there is only 1 solution.

rubikaz

I got ‘m=4’ by mental arithmetic. Because 3^m > 65 and 2^m < 3^m, that’s necessary.
The value '65' determines that the value range of m must be less than 5 and greater than 0. When m is a positive integer, test the m=5, 4, 3, 2, 1 and finally get m=4.

f-s

You can also factor 65 into 65 and 1. This gives values of a and b as 33 and 32 hence b=2^(m/2)=32, m= 10 but m will have different value for 3^(m/2)= 33. Using logs(or ln) m=(log 33/log 3)=6.365

samuelmayna

4 is the m;
My high school math teacher used to tell me, the easiest way to understand an equation is to make them look the same, that is to say, we should make the brief side more complicated other than simplifing the complicated side for the most of the time.

georiashang

Similarly you can as well re-write 65 as 81-16....From there you make the bases of the two numbers to be similar with what you have on the left hand side..From there you take one of the corresponding bases and equate them together, when bases are the same powers will also be the same.

nicholastergech

Hi there! Here is an alternative solution. The difference of the powers on the rhs increases with m, and it is already greater than 65 for m=5. So m must be less than 5. Assuming m is an integer, and noticing that m=3 doesn't work, the only possible solution is 4.

Shirlippe

Thank you very much!
Без перевода мне всë так понятно было👍

myvsepc

Thank you so much for sharing this wonderful learning for us Sir! God bless...❤❤❤

koocefb

You tried here tutor Jakes. I have learnt something here. Just keeping on running this channel. More grace, love from Port Harcourt ❤.

therichcircle.

How can one just assume that x+y = 13, and x-y = 5, respectfully, as there's 65 and 1 as well. Furthermore, this wouldn't really work if 65 had many more factors, making it more complicated, opening up to a lot more possibilities.

sanmus

Great job, You are an excellent instructor. From an admirer Ethiopian in America

hwrgbuq

Excellent explanation. It seemed very complicated, but it turned out to be easier than expected. A hug

danielmelo

A matemática é uma língua universal como a música. Parabéns, ótima técnica ❤

Sapped

Parabéns. Não sei falar nada em inglês e mesmo assim consegui aprender com sua aula. Até eu estou surpreso de ter assistido sua aula até o final sem saber se iria entender o seu modo de esnsinar. A matemática pode ser universal, mas o jeito de ensinar é fundamental.

airtonreis

Thank you for reminding me that you can't believe everything on the internet. Props to you i almost believed it untill i tested it with calculator. BRAVO YOU SHOULD WIN AN OSCAR

teamredxpro

I did it in a different way.
Write 3 as 2+1 and perform binomial expansion, so the 2^m cancels and subtract one from both sides. We have 64 on one side and some series on other side.
Notice that the series 2 + 2^2 2^m will definitely be smaller than series on left which is equal to 64.
So make this G.P. sum less than 64, you will get that m should be less than 5, and once you have known this, you have proved that you just need to find a solutions less than 5 and those will be the only solutions.

Only m=4 works out.

ravirajshelar

Math Olympiad 3^m–2^m=65 | Math Olympiad Problems | Algebra