## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$t\approx-460.517$
Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the given equation, $e^{-0.01t}=100 ,$ is \begin{array}{l}\require{cancel} \ln e^{-0.01t}=\ln100 \\\\ -0.01t(\ln e)=\ln100 \\\\ -0.01t(1)=\ln100 \\\\ -0.01t=\ln100 \\\\ t=\dfrac{\ln100}{-0.01} \\\\ t\approx-460.517 .\end{array}