Answer
$x=4$
Work Step by Step
Using the properties of logarithms, the given equation, $
\log_2 x+\log_2 (x+4)=5
$, is equivalent to
\begin{align*}\require{cancel}
\log_2 [x(x+4)]&=5
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\
\log_2 (x^2+4x)&=5
&(\text{use the Distributive Property})
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, the equation above is equivalent to
\begin{align*}\require{cancel}
x^2+4x&=2^5
\\
x^2+4x&=32
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
x^2+4x-32&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}\require{cancel}
(x+8)(x-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+8=0 & x-4=0
\\
x=-8 & x=4
.\end{array}
If $x=-8$, the term $\log_2 x$ of the original equation becomes $\log_2(-8)$. This is undefined since $\log_b x$ is defined only for $x$ and $b$ are positive real numbers.
Hence, the solution to the equation $
\log_2 x+\log_2 (x+4)=5
$ is $
x=4
$.
