## College Algebra (11th Edition)

$-2$
$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to evaluate the given expression, $\ln\left( \dfrac{1}{e^2} \right) .$ $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \ln1-\ln e^2 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \ln1-2\ln e .\end{array} Since $\ln e=1$ and $\ln1=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 0-2(1) \\\\= -2 .\end{array}