Chapter 12 - Exponential Functions and Logarithmic Functions - 12.4 Properties of Logarithmic Functions - 12.4 Exercise Set - Page 811: 89

$-2$

By definition, $\log_{a}x=2$ means $a^{2}=x.$ Then, $\log_{1/a}x=y$ means $(\displaystyle \frac{1}{a})^{y}=x$ Substituting, $(\displaystyle \frac{1}{a})^{y}=a^{2}$ $\Rightarrow\quad (a^{-1})^{y}=a^{2}$ $\Rightarrow\quad a^{-y}=a^{2}$ $\Rightarrow\quad y=-2$ $\log_{1/a}x=-2$