17. (10 points) Let the function
$$
f(x)=\log _{2}\left(a^{x}-b^{x}\right),
$$

and $f(1)=1, f(2)=\log _{2} 12$.
(1) Find the values of $a, b$;
(2) When $x \in[1,2]$, find the maximum value of $f(x)$.
Three, 17. (1) From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\log _{2}(a-b)=1, \\
\log _{2}\left(a^{2}-b^{2}\right)=\log _{2} 12
\end{array}\right. \\
\quad \Rightarrow\left\{\begin{array}{l}
a-b=2, \\
a^{2}-b^{2}=12
\end{array} \Rightarrow a=4, b=2 .\right.
\end{array}
$$
(2) From (1), we know
$$
\begin{array}{l}
f(x)=\log _{2}\left(4^{x}-2^{x}\right) \\
=\log _{2}\left[\left(2^{x}-\frac{1}{2}\right)^{2}-\frac{1}{4}\right] . \\
\text { Let } u(x)=\left(2^{x}-\frac{1}{2}\right)^{2}-\frac{1}{4} .
\end{array}
$$

By the monotonicity of composite functions, we know that $u(x)$ is an increasing function on the interval $[1,2]$.
Thus, $u(x)_{\text {max }}=\left(2^{2}-\frac{1}{2}\right)^{2}-\frac{1}{4}=12$.
Therefore, the maximum value of $f(x)$ is $\log _{2} 12=2+\log _{2} 3$.