Answer
$
f(x)=160\cdot\left(\left(1/2\right)^{1 / 40}\right)^x
$
Work Step by Step
We want to have $f(x)=a b^{x}$.
The graph shows that the points $(40,80),(120,20)$ lie on the curve of $f(x)$.
We must have
$f(40)=a b^{40}= 80$ and $f(120)=a b^{120}= 20$. This gives
$$
\begin{aligned}
\frac{a b^{120}}{a b^{40}} & =\frac{f(120)}{f(40)} \\
b^{80} & =\frac{20}{80} \\
b & =\left(1/4\right)^{1 / 80}= =\left(1/2\right)^{1 / 40} = 0.9828
\end{aligned}
$$ We determine $a$:
$$
\begin{aligned}
ab^{40}& =80\\
a\left(\left(1/2\right)^{1 / 40}\right)^{40}&=80\\
a\cdot \frac{1}{2} &= 80 \\
a & =160
\end{aligned}
$$ It follows that
$$
f(x)=160\cdot\left(\left(1/2\right)^{1 / 40}\right)^x$$
