Calculate: log_{6}(x+3)-log_{6}(x-2)=1

Expression: $\log_{6}(x+3)-\log_{6}(x-2)=1$

Add $\log_{6}(x-2)$ to both sides

$\log_{6}(x+3)-\log_{6}(x-2)+\log_{6}(x-2)=1+\log_{6}(x-2)$

Simplify

$\log_{6}(x+3)=1+\log_{6}(x-2)$

Apply log rules

$x+3=6(x-2)$

Solve $ x+3=6(x-2):{\quad}x=3$

$x=3$

The solution is

$x=3$