Answer
a) $$
V(t)= \begin{cases}80 & \text { for } t<0 \\ 80(0.8459)^t & \text { for } t \geq 0\end{cases}
$$
b) $t\approx 39.94$
Work Step by Step
a) Let $v(t)=a(b)^t $ for the exponential part of the voltage as shown in the figure. The value of $a$ can be read directly from the figure as $a= 80$. We now have $v(t)= 80(b)^t$. Use the point (10,15) on the graph to find the value of $b$.
$$
\begin{aligned}
v(10) & = 15 \\
80b^{10} & =15 \\
b^{10} & =\frac{15}{80} \\
b & =0.1875^{1 / 10} \\
& =0.8459
\end{aligned}
$$
$v(t)=80(0.8459)^t$
For $t<0$, we have $V=80$. The final expression for the voltage is
$$
V(t)= \begin{cases}80 & \text { for } t<0 \\ 80(0.8459)^t & \text { for } t \geq 0\end{cases}
$$
b) We determine $t$ so that $V(t)=0.1$:
$$\begin{aligned}
80(0.8459)^t&=0.1\\
(0.8459)^t&=\frac{0.1}{80}\\
(0.8459)^t&=0.00125\\
t\ln 0.8459&=\ln 0.00125\\
t&=\frac{\ln 0.00125}{\ln 0.8459}\\
&\approx 39.94.
\end{aligned}$$
