Let p = 2^x

Then, 1^x+2^x =4^x

Gives: p+1 =p^2

--> p^2-p-1=0

--> p = (1+/-v5)/2

However, we note that 1-v5<0, and thus we must disregard the root p = (1-v5)/2, since log_2(p) will then turn out complex or imaginary, which is not deemed as irrational.

Therefore, p = (1+v5)/2 =1.6180339887

Then, we must have: x = log_2(p) =log(1.6180339887)/log(2)

[/EDIT]=~ 0.6942419136

*Edited on ***March 31, 2023, 10:40 pm**