2^3x-1=(1\8)^x

If this is

$$2^{3x-1}=(\frac{1}{8})^x$$

then rewrite it as

$$\\2^{3x-1}=(\frac{1}{2^3})^x\\\\2^{3x-1}=(2^{-3})^x\\\\2^{3x-1}=2^{-3x}$$

So we must have

$$\\3x-1=-3x\\\\6x=1\\\\x=\frac{1}{6}$$

.

x=8^(-x-1)*((8^x)+1)