Answer
$
f(x)=\left(8 \cdot d^{-0.5}\right)\cdot\left( \frac{d^{0.5}}{2}\right)^x
$
Work Step by Step
We want to find $f(x)=ab^x$, given $f(1)=4$ and $f(3)=d$. We now find $b$ and $a$.
$$
a b^1=4 \quad \text { and } \quad a b^3=d
$$ $$
\begin{aligned}
&\frac{a b^3}{a b^1}=\frac{d}{4}\\
&\begin{aligned}
b^2 & =\frac{d}{4} \\
b & =\frac{d^{0.5}}{2}
\end{aligned}
\end{aligned}
$$ and $$
a b^1=4=f(1)
$$ $$
\begin{aligned}
a\left(\frac{d^{0.5}}{2}\right) & =4 \\
a & =\frac{8}{d^{0.5}}
\end{aligned}
$$ $$
\begin{aligned}
f(x) & =a b^x \\
& =\left(\frac{8}{d^{0.5}}\right)\left(\frac{d^{0.5}}{2}\right)^x \\
&= \left(8 \cdot d^{-0.5}\right)\cdot\left( \frac{d^{0.5}}{2}\right)^x
\end{aligned}
$$ Hence $$
f(x)=\left(8 \cdot d^{-0.5}\right)\cdot\left( \frac{d^{0.5}}{2}\right)^x
$$
