[Algebra]
Problem: Let $f(x)=2^{k x}+9$, where $k$ is a real number. If $f(3): f(6)=1: 3$, determine the value of $f(9)-f(3)$.

Solution: From the given condition,

$$
\begin{aligned}
\frac{f(3)}{f(6)}=\frac{2^{3 k}+9}{2^{6 k}+9} & =\frac{1}{3} \\
3\left(2^{3 k}+9\right) & =2^{6 k}+9 \\
0 & =2^{6 k}-3\left(2^{3 k}\right)-18 .
\end{aligned}
$$

We treat this as a quadratic equation in the variable $x=2^{3 k}$, so

$$
\begin{aligned}
& 0=x^{2}-3 x-18 \\
& 0=(x-6)(x+3)
\end{aligned}
$$

Therefore, $2^{3 k}=6$ or $2^{3 k}=-3$. Since $2^{a}>0$ for any $a$, then $2^{3 k} \neq-3$.

So $2^{3 k}=6$. We could solve for $k$ here, but this is unnecessary.



We calculate $f(9)-f(3)=\left(2^{9 k}+9\right)-\left(2^{3 k}+9\right)$

$$
\begin{aligned}
& =2^{9 k}-2^{3 k} \\
& =\left(2^{3 k}\right)^{3}-2^{3 k} \\
& =6^{3}-6 \\
& =210 .
\end{aligned}
$$

Therefore $f(9)-f(3)=210$.

Final answer: 210