## College Algebra (11th Edition)

$t=e^{\frac{p-r}{k}}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $r=p-k\ln t ,$ for $t ,$ use the properties of equality to isolate the needed variable. Then change to exponential form. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate $t ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} k\ln t=p-r \\\\ \dfrac{k\ln t}{k}=\dfrac{p-r}{k} \\\\ \ln t=\dfrac{p-r}{k} .\end{array} Since $\ln x=\log_e x$, the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e t=\dfrac{p-r}{k} .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} t=e^{\frac{p-r}{k}} .\end{array}