#### Answer

$x=12$

#### Work Step by Step

Using the Quotient Property of logarithms, the given expression, $
\log_2(x^2-100)-\log_2(x+10)=1
$ is equivalent to
\begin{array}{l}\require{cancel}
\log_2\dfrac{x^2-100}{x+10}=1
.\end{array}
Changing to exponential form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{x^2-100}{x+10}=2^1
\\\\
\dfrac{x^2-100}{x+10}=2
.\end{array}
Using the properties of equality and the concepts of quadratic equations, the solution/s to the equation above is/are
\begin{array}{l}\require{cancel}
x^2-100=2(x+10)
\\\\
x^2-100=2x+20
\\\\
x^2-2x-100-20=0
\\\\
x^2-2x-120=0
\\\\
(x-12)(x+10)=0
.\end{array}
Equating each factor to zero and then solving for the variable, the solutions are $
x=\left\{ -10,12 \right\}
.$
Upon checking, only $
x=12
$ satisfies the original equation.