## College Algebra (11th Edition)

$t=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right)$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $I=\dfrac{E}{R}(1-e^{-Rt/2}) ,$ in terms of $t ,$ use the properties of equality and logarithms to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{R}{E}\cdot I=\dfrac{R}{E}\cdot\dfrac{E}{R}(1-e^{-Rt/2}) \\\\ \dfrac{RI}{E}=1-e^{-Rt/2} \\\\ e^{-Rt/2}=1-\dfrac{RI}{E} .\end{array} Taking the natural logarithm of both sides results to \begin{array}{l}\require{cancel} \ln e^{-Rt/2}=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} -\dfrac{Rt}{2}\ln e=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Since $\ln e=1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -\dfrac{Rt}{2}(1)=\ln \left(1-\dfrac{RI}{E} \right) \\\\ -\dfrac{Rt}{2}=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Using the properties of equality to isolate the needed variable results to \begin{array}{l}\require{cancel} -\dfrac{2}{R}\left(-\dfrac{Rt}{2}\right)=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right) \\\\ t=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right) .\end{array}