Solve 3log1^2-2log1^3=1log1(1/8)

Answer 1

I have no idea what log1^x is supposed to mean
log1 = 0 for any base, but log(base 1) makes no sense at all.
You have unusual notation, but I'll go with

3log2 - 2log3 = log(1/8)
3log2 - 2log3 = -3log2
6log2 - 2log3 = 0
log(2^6/3^2) = 0
2^6/3^2 = 1
which is not true

do you need some variables somewhere? Most equations do.
Try typing log_b(exp) for log(base b) exp