## College Algebra (11th Edition)

$x=e^{\frac{k}{p-a}}$
$\bf{\text{Solution Outline:}}$ 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}