Chapter 4 - Section 4.3 - Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions - Page 427: 42

$x=\dfrac{1}{9}$

Since $y=b^x$ is equivalent to $\log_b y=x,$ then the given equation, $ \log_3x=-2 ,$ is equivalent to \begin{array}{l}\require{cancel} 3^{-2}=x .\end{array} Using $a^{-m} = \frac{1}{a^m}, a\ne0$, gives \begin{array}{l}\require{cancel} \dfrac{1}{3^2}=x \\\\ \dfrac{1}{9}=x \\\\ x=\dfrac{1}{9} .\end{array}