## College Algebra (11th Edition)

$x=12$
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.