## College Algebra (11th Edition)

$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $2^x=\log_2 16 ,$ use the properties of logarithms to simplify the right side. Then express both sides in the same base and equate the exponents. $\bf{\text{Solution Details:}}$ Using exponents, the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=\log_2 2^4 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} 2^x=4\log_2 2 .\end{array} Since $\log_b b =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=4(1) \\\\ 2^x=4 \\\\ 2^x=2^2 .\end{array} Since the bases are the same, then the exponents can be equated. That is, \begin{array}{l}\require{cancel} x=2 .\end{array}