Answer
$t=e^{\frac{p-r}{k}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}