Answer
$x=e^{\frac{k}{p-a}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
p=a+\dfrac{k}{\ln x}
,$ for $
x
,$ 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 $
x
,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
p-a=\dfrac{k}{\ln x}
\\\\
\ln x(p-a)=\left( \dfrac{k}{\ln x} \right) \ln x
\\\\
\ln x(p-a)=k
\\\\
\ln x=\dfrac{k}{p-a}
.\end{array}
Since $\ln x=\log_e x$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\log_e x=\dfrac{k}{p-a}
.\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}
x=e^{\frac{k}{p-a}}
.\end{array}