Answer
$x=8$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log_2 x+\log_2(x-7)=3
$, is equivalent to
\begin{align*}\require{cancel}
\log_2 [x(x-7)]&=3
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\
\log_2 (x^2-7x)&=3
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, then the equation above implies
\begin{align*}\require{cancel}
x^2-7x&=2^3
\\
x^2-7x&=8
\\
x^2-7x-8&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}\require{cancel}
(x-8)(x+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions of the equation above are
\begin{array}{l|r}
x-8=0 & x+1=0
\\
x=8 & x=-1
.\end{array}
If $x=-1$, the term $\log_2x$ becomes undefined since $\log_bx$ is defined only when $b$ and $x$ are nonzero. Thus, only $
x=8
$ is the acceptable solution.
Hence, the solution to the equation $
\log_2 x+\log_2(x-7)=3
$ is $
x=8
$.
