## Precalculus: Mathematics for Calculus, 7th Edition

$x=3$
$\log_{8}(x+5)-\log_{8}(x-2)=1$ Combine the subtraction on the left side as the logarithm of a division: $\log_{8}\dfrac{x+5}{x-2}=1$ Rewrite this equation in exponential form: $\dfrac{x+5}{x-2}=8^{1}$ $\dfrac{x+5}{x-2}=8$ Solve for $x$: $x+5=8(x-2)$ $x+5=8x-16$ $x-8x=-16-5$ $-7x=-21$ $x=\dfrac{-21}{-7}$ $x=3$