Chapter 12 - Exponential Functions and Logarithmic Functions - 12.3 Logarithmic Functions - 12.3 Exercise Set - Page 804: 109

$-2$

For $x\gt 0$ and $a$ a positive constant other than 1, $\log_{a}x$ is the exponent to which $a$ must be raised in order to get $x$. Thus, $\log_{a}x=m$ means $a^{m}=x$ --- By definition, $\log_{1/5}25=x$ means $x$ is the exponent with which we raise ($\displaystyle \frac{1}{5}$) to obtain $25.$ $5^{2}=25$ $(5^{-1})^{-2}=25$ $(\displaystyle \frac{1}{5})^{-2}=25$ so, $x=-2$ $\log_{1/5}25=-2$