1231

${\displaystyle f(x)}$ 的定义域为 ${\displaystyle (1,+\mathrm{\infty })}$, 在此之上 ${\displaystyle f^{\prime }(x)=a^{x-1}\ln a-\frac{1}{(x-1)\ln a}}$ 是一个保号函数.
当 ${\displaystyle a>1}$, ${\displaystyle \underset{x\to 1^{+}}{lim}{f}^{\prime }(x)=\ln a-\underset{x\to 1^{+}}{lim}\frac{1}{(x-1)\ln a}=-\mathrm{\infty }}$, 但 ${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{f}^{\prime }(x)=\underset{x\to +\mathrm{\infty }}{lim}{a}^{x-1}\ln a=+\mathrm{\infty }}$, 因此发生了变号. 因此 ${\displaystyle 0<a<1}$.
此时根据 ${\displaystyle \underset{x\to 1^{+}}{lim}{f}^{\prime }(x)}$ 的符号确定 ${\displaystyle f^{\prime }(x)\ge 0,\mathrm{\forall }x>1}$.这个不等式等价于 $$(x-1)a^{x-1}\le \frac{1}{{\ln }^{2}a}.$$ 记左边函数为 ${\displaystyle g(t)=g(x-1)}$, 则 ${\displaystyle g^{\prime }(t)=a^t(1+t\ln a)}$, 因此 ${\displaystyle g(t)}$ 有极大值点 ${\displaystyle -\frac{1}{\ln a}}$, 记为 ${\displaystyle -b}$, 因此只需 $$\begin{array}{rl}g(-b)=-b{a}^{-b}\le b^2⟺& -{\mathrm{e}}^{-b\ln a}=-{\mathrm{e}}^{-1}\ge b,\end{array}$$ 从而 ${\displaystyle a={\mathrm{e}}^{b^{-1}}\ge {\mathrm{e}}^{-\mathrm{e}}}$. 综上, ${\displaystyle a\in [{\mathrm{e}}^{-\mathrm{e}},1)}$.