Chapter 4 - Section 4.3 - Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions - Page 427: 24

$\log_{4} \dfrac{1}{8}=-\dfrac{3}{2}$

Since $y=b^x$ is equivalent to $\log_b y=x,$ then the given equation, $ 4^{-3/2}=\dfrac{1}{8} ,$ is equivalent to \begin{array}{l}\require{cancel} \log_{4}{(\frac{1}{8})}=-\frac{3}{2} .\end{array}