Answer
The solution set is $\{4\}$.
Work Step by Step
The given equation is
$\log _6(x+5) +\log _6x=2$
Use product rule.
$\log _6 [(x+5)x]=2$
Rewrite in exponential form.
$(x+5)x= 6^2$
Use the distributive property on the left.
$x^2+5x=36$
Subtract $36$ from both sides.
$x^2+5x-36=36-36$
Simplify.
$x^2+5x-36=0$
Factor.
$x^2+9x-4x-36=0$
$x(x+9)-4(x+9)=0$
$(x+9)(x-4)=0$
Set both factors equal to $0$.
$x+9=0$ or $x-4=0$
Isolate $x$.
$x=-9$ or $x=4$
$-9$ is not in the domain of a logarithmic function.
The solution is $x=4$.
