Answer
$t=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
