College Algebra, Chapter 5, 5.2, Section 5.2, Problem 86

If the rate at which a battery charges is slower the closer the battery is to its maximum charge $C_0$, then the time (in hours) required to charge a fully discharged battery to a charge $C$ is given by

$\displaystyle t = - k \ln \left( 1 - \frac{C}{C_0} \right)$

where $k$ is a positive constant that depends on the battery. For a certain battery, $k = 0.25$. Suppose this battery is fully discharged, how long will it take to charge to $90 \%$ of its maximum charge $C_0$?

If charge $C$ is $90 \%$ of charge $C_0$, then

$C = 0.9 C_0$

So,


$
\begin{equation}
\begin{aligned}

t =& - k \ln \left( 1 - \frac{0.9 \cancel{C_0}}{\cancel{C_0}} \right)
&& \text{Substitute } C = 0.9 C_0
\\
\\
t =& -0.25 \ln (1 - 0.9)
&& \text{Simplify}
\\
\\
t =& 0.58 \text{ hours}

\end{aligned}
\end{equation}
$


This means that it will take $0.58$ hours to charge to $90 \%$ of its maximum charge $C_0$.