Question

Rewrite problem in equivalent logarithmic form.
1/2=32^-1/5

Answer 1

so write 1/2 as 2^{-1} then

\[2^{-1} = 32^{-\frac{1}{5}}\]

take the log of both sides and remember the log law for powers

\[\log(x^a) = a \times \log(x)\]

apply this law to the problem