## Calculus with Applications (10th Edition)

Definition of Logarithm: For $a > 0, a\neq 1$, and $x > 0$, $y=\log_{a}x$ means $a^{y}=x$. $....................................$ Here, $a=3 > 0, y=-2,$and $x=\displaystyle \frac{1}{9},$ so $3^{-2}=\displaystyle \frac{1}{9}$ ( $a^{y}=x)$ means $\displaystyle \log_{3}(\frac{1}{9})=-2$ $(y=\log_{a}x).$ The statement is true.