Answer
$\left\{4\right\}$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log_4 x+\log_4 (8-x)=2
$, is equivalent to
\begin{align*}\require{cancel}
\log_4 [x(8-x)]&=2
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\
\log_4 (8x-x^2)&=2
&(\text{use the Distributive Property})
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, the equation above implies
\begin{align*}\require{cancel}
8x-x^2&=4^2
\\
8x-x^2&=16
\\
0&=x^2-8x+16
\\
x^2-8x+16&=0
.\end{align*}
Using the factoring of trinomials, the equation above is equivalent to
\begin{align*}\require{cancel}
(x-4)(x-4)&=0
\\
(x-4)^2&=0
.\end{align*}
Taking the square root of both sides (Square Root Property), and solving the variable, then
\begin{align*}\require{cancel}
x-4&=\pm\sqrt{0}
\\
x-4&=0
\\
x-4+4&=0+4
\\
x&=4
.\end{align*}
Hence, the solution set of the equation $
\log_4 x+\log_4 (8-x)=2
$ is $
\left\{4\right\}
$.
