Answer
$\left\{3\right\}$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log_8(x+5)+\log_8(x-2)=1
,$ is equivalent to
\begin{align*}\require{cancel}
\log_8 [(x+5)(x-2)]&=1
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, the equation above implies
\begin{align*}\require{cancel}
(x+5)(x-2)&=8^1
\\
(x+5)(x-2)&=8
.\end{align*}
In the form $ax^2+bx+c=0$, the equation above is equivalent to
\begin{align*}\require{cancel}
x(x)+x(-2)+5(x)+5(-2)&=8
&(\text{use }(a+b)(c+d)=ac+ad+bc+bd)
\\
x^2-2x+5x-10&=8
\\
x^2+(-2x+5x)+(-10-8)&=0
\\
x^2+3x-18&=0
.\end{align*}
Using the factoring of trinomials, the equation above is equivalent to
\begin{align*}\require{cancel}
(x+6)(x-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+6=0 & x-3=0
\\
x=-6 & x=3
.\end{array}
If $x=-6$, the term $\log_8 (x+5)$ of the original equation becomes $\log_8 (-1)$. This is undefined since in $\log_b x$, $x$ and $b$ should be positive numbers with $b\ne1$.
Hence, the solution set of the equation $
\log_8(x+5)+\log_8(x-2)=1
$ is $
\left\{3\right\}
$.
