Answer
$t\approx \left(\dfrac{25}{10.7}\right)^{3.7}$
Year $2018$
Work Step by Step
We are given the function:
$$P=10.7t^{0.272}.$$
Find the inverse of the function:
$$\begin{align*}
P&=10.7t^{0.272}\quad&&\text{Write original function.}\\
\dfrac{P}{10.7}&=t^{0.272}\quad&&\text{Divide each side by }10.7.\\
\left(\dfrac{P}{10.7}\right)^{1/0.272}&=(t^{0.272})^{1/0.272}\quad&&\text{Raise each side to power }\dfrac{1}{0.272}.\\
\left(\dfrac{P}{10.7}\right)^{3.7}&\approx t\quad&&\text{Simplify. }
\end{align*}$$
The inverse of $P$ is $t\approx \left(\dfrac{P}{10.7}\right)^{3.7}$.
Determine $t$ for $P=25$:
$$t=\left(\dfrac{25}{10.7}\right)^{3.7}\approx 23.$$
The year will be $1995+23=2018$.
