The equation is equivalent to log (base 2 ) (X-1)^3 (4) = 5 , therefore ,
log(base 2)[X^3-3X^2+3X-1](4) = 5 , so , 2^5 = 4X^3-12X^2+12X-4 , we simplify a little bit more , then 32 = 4(X^3 - 3X^2 + 3X - 1) , then : 8 = X^3 - 3X^2 + 3X - 1 , putting the whole thing in standard form yuo get X^3 - 3X^2 + 3X - 9 = 0 , which you factorize in this way : X^2(X-3) 3(X-3) = 0 , (X-3)(X^2 +3) =0 , implying that X = 3 , X = 3^(1/2) i and X = -3^ (1/2)i three roots because is a third degree equation . Because we were asked to express the irrational answer rounded up to the third place , then we add : 3^(1/2) i = 1.732 i and its pair ( every irrational root comes in pairs ) would be -1.732 i