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[topquark]

It is so painful and slow to watch you do this in such minute detail, without effecting obvious simplifications immediately. I solved it in my head in about 30 seconds.

[walterwen]

Math Olympiad Problem: 9^(4^m) = 4^(9^m); m = ?
log[9^(4^m)] = log[4^(9^m)], (9^m)log4 = (4^m)log9
(9^m)/(4^m) = log₄9 = 1.585; (9/4)^m = 1.585, 2.25^m = 1.585
log(2.25^m) = log1.585, m = (log1.585)/(log2.25) = 0.200/0.352 = 0.568
Answer check:
9^(4^m) = 9^(4^0.568) = 125.068
4^(9^m) = 4^(9^0.568) = 125.097= 9^(4^m); Confirmed
The calculation was achieved on a smartphone with a standard calculator app
Final answer:
m = 0.568

[elgone]

Me he perdido en este paso: 2m=log(log2/3)/log2/3, que pasa a 2m=log(2/3)log(3)2 y finaliza con m=1/2log(2/3)log(3)2. Ruego aclaración.

[OgunbodedeTunde-cg]

Teach me more on log because I'm going for a math competition you might even coach me.what do you think?

[АндрейПергаев-зн]

А почему нельзя сразу логарифмировать по базе 4 или 9?

[emanueltavares]

Use ln instead of log, and the solution comes simpler and faster.

[tunneloflight]

trivial solution: m = 0. (9^4)^m=1 ergo m=0, (9^4)^0 = 1, (4^9)^0=1 check. 1 = 1. Also 9^(4*0)=9^0=1, 4^(9*0)=4^0=1. check. for non-zero m, (9^4) must equal (4^9). But that is nonsense. For m=0, both sides are 1.

[ludmilak]

а^m ??? q^m??? 9^m???
Научитесь писать буквы и цифры!

[edmundodejesusmarcanorojas]

This root was missing: -27.2436654839064