Five. (Full marks 20 points) Let the constant $a>1>b>0$. Then, under what relationship between $a$ and $b$ is the solution set of $\lg \left(a^{x}-b^{x}\right)>0$ $(1, +\infty)$?
Five, $\because a^{\prime}-b^{\prime}>0$, i.e., $\left(\frac{a}{\dot{v}}\right)^{\prime}>1$, and $\frac{a}{b}>1$,
$\therefore$ when $x \in(0,+\infty)$, to make the solution set of the inequality $(1,+\infty)$, it is only necessary that $f(x)=\lg \left(a^{x}-b^{x}\right)$ is an increasing function on $(0,+\infty)$, and $f(1)=0$.
Let $x_{2}>x_{1}>0$. If
$$
\begin{aligned}
f\left(x_{1}\right)-f\left(x_{2}\right) & =\lg \left(a^{x_{1}}-b^{x_{1}}\right)-\lg \left(a^{x_{2}}-b^{x_{2}}\right) \\
& =\lg \left(\frac{a^{x_{1}}-b^{x_{1}}}{a^{x_{2}}-b^{x_{2}}}\right)x_{1}>0, a>1>b>0, \\
\therefore a^{x_{1}}-a^{x_{2}}0 .
\end{aligned}
$$

Thus, equation (*) holds.
$\therefore f(x)=\lg \left(a^{x}-b^{x}\right)$ is an increasing function on $(0,+\infty)$.
Also, $f(1)=\lg (a-b)=0$, so $a-b=1, a=b+1$.