Can you show your work, please? Thanks.
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Answer:
t = (L/R)ln(V/(V -iR))
Step-by-step explanation:
Undo what is done to t. The exponential is undone using logarithms.
[tex]i=\dfrac{V}{R}\left[1-e^{-(R/L)t}\right]\qquad\text{given}\\\\\dfrac{iR}{V}=1-e^{-(R/L)t}\qquad\text{multiply by $R/V$}\\\\e^{-(R/L)t}=1-\dfrac{iR}{V}\qquad\text{add $e^{( )}-iR/V$}\\\\-(R/L)t=\ln{\left[1-\dfrac{iR}{V}\right]}\qquad\text{take natural log}\\\\t=-\dfrac{L}{R}\ln{\left[1-\dfrac{iR}{V}\right]}\qquad\text{divide by the coefficient of t}\\\\\boxed{t=\dfrac{L}{R}\ln\left[\dfrac{V}{V-iR}\right]}\qquad\text{eliminate leading minus sign}[/tex]