#### Answer

$x=0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log_8(x+2)+\log_8(x+4)=\log_88
,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use the properties of equality and concepts on solving quadratic equations. Finally, do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_8[(x+2)(x+4)]=\log_88 .\end{array}
Since the logarithm on both sides have the same base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (x+2)(x+4)=8 .\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x)+x(4)+2(x)+2(4)=8 \\\\ x^2+4x+2x+8=8 \\\\ x^2+6x+8=8 .\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2+6x+8-8=0 \\\\ x^2+6x=0 .\end{array}
Factoring the $GCF=x,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
x(x+6)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+6=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+6=0
\\\\
x=-6
.\end{array}
If $
x=-6
,$ the part of the given expression, $
\log_8(x+2)
,$ becomes $
\log_8(-4)
.$ This is not allowed since $
\log x$ is defined only for positive values of $x.$ Hence, only $
x=0
$ satisfies the original equation.