A Very Nice Math Olympiad Problem | Solve for x? | Algebra Equation

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.
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[@zihaoooi787]

attempting this before the video starts
8^log(x) - 2^log(x) = 5!
5! is trivially 120.
8^log(x) - 2^log(x) = 120
(2^3)^log(x) - 2^log(x) = 120
(2^log(x))^3 - 2^log(x) = 120
let u = 2^log(x)
u^3 - u = 120
u(u^2 - 1) = 120
(u-1)u(u+1) = 120
after factorizing 120 and using the simple process of using your eyes, we get 4 * 5 * 6 = 120.
therefore u = 5
2^log(x) = 5
log(x) = log_2(5)
x = 10^log_2(5) approx. 209.85924

[@pspprabhat]

=>(2^log×)^3-2^log×
=120=5^3-5
=>2^logx=5
&logx=log@2 (5)
=>×=10^log@2(5)