A Very Nice Math Olympiad Problem | Can You Solve for x? | Algebra | Exponential

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.
Please feel free to share your ideas in the comment section.
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[@1234larry1]

Although most would simplify to base 2 for the answer, it would be easier to stay in base 4, so that in checking the answer, you eliminate another step. Also, you could have started out when you have (x-2)log4=log3+log5 by simply dividing both sides by log4, eliminating another step.

[@thetsukad748]

I don't think, you should make it so complicated. Use the basic logarithmic identity
a^log_a(b) = b
4^(x-2) = 15
4^(x-2) = 4^log_4(15)
Bases are equal, therefore powers are also equal
x - 2 = log_4(15)
x = log_4(15) + 2

[@rdfernandes13]

or you could continue calling 2 = log_4(4^2)
then
x = log_4(15) + log_4(16)
x = log_4(240) , then you have the smallest version of the answer

[@rdfernandes13]

I got to this answer using some others basics
4^(x-2) = 15 ; log both sides
(x-2)log4 = log15
x - 2 = log15/log4 = log_4(15) ; log division let us use divisor as base
x = log_4(15) + 2
and then keep as I have before
calling 2 = log_4(4^2)
then
x = log_4(15) + log_4(16)
x = log_4(240)

[@key_board_x]

4^(x - 2) = 15
Ln[4^(x - 2)] = Ln(15)
(x - 2).Ln(4) = Ln(15)
x - 2 = Ln(15) / Ln(4)
x = 2 + [Ln(15) / Ln(4)]
x = [2.Ln(4) + Ln(15)] / Ln(4) → to go further
x = [Ln(4²) + Ln(15)] / Ln(4)
x = Ln(4² * 15) / Ln(4)
x = Ln(240) / Ln(4)

[@enisogdum1812]

We are just laughing like these questions

[@prollysine]

4^(x-2)=4^(ln15/ln4) , x-2=ln15/ln4 , x=ln15/ln4+2 , test , 4^(ln15/ln4+2-2)=15 , OK ,