Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 826: 96

$-3$

In the parentheses, $\log_{125}5=A$ means, by definition, that $125^{A}=5$ $125=5^{3}$, so $(5^{3})^{A}=5\displaystyle \quad\Rightarrow 5^{3A}=5^{1}\quad\Rightarrow\quad A=\frac{1}{3}$ $\log_{5}125=3$, because $5^{3}=125$ So, we have$\displaystyle \quad x=(\frac{1}{3})^{3}=3^{-3}$ Finally, $\log_{3}x=\log_{3}3^{-3}=-3$