## Intermediate Algebra (6th Edition)

$x=\frac{9}{5}$
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $log_{7}(5x-2)=1$ is equivalent to $7^{1}=7=5x-2$. For $7=5x-2$, add 2 to both sides. $5x=9$ Divide both sides by 5. $x=\frac{9}{5}$