Answer
$
f(x)=50\left(\sqrt[5]{\frac{2}{5}}\right)^x\approx 50\cdot\left(0.8326\right)^x
$
Work Step by Step
We want to have $f(x)=a b^{x}$.
The graph shows that the points $(0,50),(5,20)$ lie on the curve of $f(x)$. We must have $f(0)=a b^{0}=a= 50$ and so $f(x)=50 b^{x}$.
But $f(5)=50 b^{5}= 20\implies b= 0.4^{1/5}=\sqrt[5]{\frac{2}{5}}\approx 0.8326 $
It follows that $$
f(x)=50\left(\sqrt[5]{\frac{2}{5}}\right)^x\approx 50\cdot\left(0.8326\right)^x
$$
