# Chapter 9 - Sections 9.1-9.6 - Integrated Review - Functions and Properties of Logarithms: 23

$x=3$

#### Work Step by Step

We are given that $25^{x}=125^{x-1}$. Both of these numbers are powers of 5, so we can rewrite the equation as $(5^{2})^{x}=(5^{3})^{x-1}$. This can be rewritten as $5^{2x}=5^{3x-3}$ From the uniqueness of $b^{x}$, we know that $b^{x}=b^{y}$ is equivalent to $x=y$ (when $b\gt0$ and $b\ne1$). Therefore, $2x=3x-3$. To solve for x, subtract 3x from both sides. $-x=-3$ Divide both sides by -1. $x=3$

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