How To Solve 2^m–2^n=8064 | Math Olympiad | Algebra 1 | Math Challenge.

How To Solve 2^m–2^n=8064 for m and n step-by-step with easy.
In solving this Math Olympiad question, I will guide you systematically on how to solve two unknown variables when they are both at the exponents with the applications of some mathematical rules without missing any step.
Here, I will equally lead you on how to apply the laws of indices and the product rule all together in solving this Math Olympiad problem/challenge.

Do help to get this video to as many that need it across the globe by sharing it via the link bellow:

You can equally watch previous videos on this channel visa the channel link bellow too:
channel/UCQvzmvJzwbhNfZqLc9RvwmA

Above all, drive more traffics and more subscribers to your channel using promoterkit through the link here:

Love you all...❤️❤️💕💕💖💖😍🙋

#howto #math #matholympiad #unknown #exponents #exponential #matholympiadquestion #matholympiadtraining #matholympiadpreparation #solution #mathstriks #mathskills #howtomakemoneyonline

[maths]

Since you didn't say that m and n must be integers, technically there are infinitely many solutions for m and n in the form of n=log2(8064/(2^k-1)); m=log2(8064/(2^k-1)+8064) where k can be any number except 0. If you plug in k=6 you get the integer solution for m and n.

[conordoran]

I watch tons of maths videos, but I never comment. I liked the approach here a lot. I didn’t know how to solve it algebraically initially but your approach really gave me that “light bulb” moment that all maths people love. Great stuff!

[JPTaquari]

In this case, the case of this problem, it's easy to solve mentally, because anyone knows that 2^10 = 1024, so just double it until you reach 8, 192 ( 2^13) .
There it is simple, because 8192 - 128 = 8064
128 = 2^6

But for a higher base, then it starts to complicate, so let's use herr professor's method, which is simple and friendly !!!!

Bingo, from Brazil

[danielfranca]

The first explanation gives an insight into the solutions. Thanks for this solution Jakes.

[andrewlayton]

Although you did not specify m, n as integers, I assumed that was the intent. The first statement in my solution was 2^m > 8064 which leads to m >= 13 quickly. Different steps, same solution.

[adryelbarros]

My solution:
In binary, 2ⁿ is represented as 1000... Witn n zeroes, so the diference 2^m - 2ⁿ, in binary, must be
11..11000..00, a number with m digits and n trailing zeroes.

Since

8064 =
We have m=13 (13 digits) and n=7 (7 zeroes at the end)

[abdesselambennour]

Merci professeur je vous suis très bien salut from morroco

[emmanuelnweke]

What are the conditions for choosing the products of 8064, because there are may of the products like 32and252, 64and126 etc

[shoshosalah]

Exactly my solution and my steps . Thank u for ur fantastic content ❤🎉

[jpbohemond]

ok but how do you find that 8064 equal to 128x63 ?

[iKF]

How do we know or predict the breaking down of 8064 should be 128 and 63? There are so many combination. Thank you.

[jerrypaquette]

I don't understand the step where you come up with 2^n(2^k-1) =128 -63.
A much better approach would be to think about the powers of 2:
1, 2, 4 1024, 2048, 4096, 8192. 1024 = 2^10. 8192 = 2 ^13, 8192 - 8064 = 128 = 2^7.
Therefore m = 13, and n =7.

[kientrucnguyen]

Chia hai vế cho 2^n sẽ ra hai bên đều lẽ- suy ra n = 7

[rekiaouhaji]

يمكن بناء اي مفاعل نووي بنفس المساحة و الوزن كدالك .،

[syedmdabid]

Facilis! m = 2, n=1 responsi.
Value of W- n(2) =???? ( valum numericus=????)

[桜木秋水]

8064
=8192 - 128
=2^13 - 2^7

hence, m=13, n=7

[oahuhawaii]

2^m - 2^n = 8064
2^m = 2^n + 8064
m = log(2^n + 8064)/log(2), for n real.

If m and n are integers, then the solution requires more work:
2^m - 2^n = 8064 = 128*63 = 2^7*(2^6 - 2^0) = 2^13 - 2^7
m = 13, n = 7

[НиколайЧуприк-ъс]

I see 8192–128=8064 from the beginning.

[rekiaouhaji]

المتر و النيوتون متر بينهما مساحة و هي غير معيرة بالاسم و ما قبل متر و ما بعد نيومتر و كيف تسلمت علم المساحة بحساب العد الزمكوني ام بقياس نيتشه

[charlesmitchell]

How did you know to factor 8064 with 128 and 63?