#### Answer

$x=2$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log_2 x+ \log_2(x+2)=3
,$ use the properties of logarithms to simplify the expression at the left. Then convert to exponential form and use the concepts of solving quadratic equations. Do checking of solution/s 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 equation above is equivalent
\begin{array}{l}\require{cancel}
\log_2 [x(x+2)]=3
.\end{array}
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
x(x+2)=2^3
\\\\
x(x+2)=8
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(x)+x(2)=8
\\\\
x^2+2x=8
\\\\
x^2+2x-8=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(x+4)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
x+4=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+4=0
\\\\
x=-4
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
If $x=-4,$ the part of the given expression ,$
\log_2 x
,$ becomes $
\log_2 -4
,$ which is not a real number. Hence, oly $
x=2
,$ satisfies the original equation.