College Algebra 7th Edition

$x=3$
We solve: $\log_{8}(x+5)-\log_{8}(x-2)=1$ $\log_{8}\frac{x+5}{x-2}=1$ $8^1=\frac{x+5}{x-2}$ $x+5=8(x-2)$ $x+5=8x-16$ $7x=21$ $x=3$