Michael is driving a car in a cold winter day with a temperature of 20 degree F and the engine of the car will overheat at about 220 degree F and if the car is parked the engine will cool down. Suppose that the equation $ln \left( \frac{T-20}{200} \right) = -0.11t$ where $T$ is the temperature of the engine and $t$ is the minutes after you park.
a.) Solve the equation for $T$.
b.) Determine the temperature of the engine after 20 min (t = 20) by using your answer in part(a).
a.)
$
\begin{equation}
\begin{aligned}
\ln \left( \frac{T - 20}{200} \right) =& -0.11 t
\\
\\
e^{\ln \left( \frac{T - 20}{200} \right)} =& e^{-0.11 t}
\\
\\
\frac{T - 20}{200} =& e^{-0.11 t}
\\
\\
T - 20 =& 200 e^{-0.11 t}
\\
\\
T =& 20 + 200 e^{-0.11 t}
\end{aligned}
\end{equation}
$
b.) if $t = 20 $ mins, then
$
\begin{equation}
\begin{aligned}
T =& 20 + 200 e^{-0.11(20)}
\\
\\
T =& 42.16^{\circ} F
\end{aligned}
\end{equation}
$
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