When $x=2$, $\varphi(a)=\mathrm{e}^{2}-a \ln a+a, \varphi^{\prime}(a)=-\ln a, \varphi(a)$ is monotonically increasing on $(0,1)$ and monotonically decreasing on $(1,+\infty)$, and $\varphi(0)=\mathrm{e}^{2}, \varphi\left(\mathrm{e}^{2}\right)=0$, thus $f(2)>0 \Rightarrow 00 \Rightarrow f^{\prime}(x)$ is monotonically increasing on $(1,+\infty)$, and as $x \rightarrow 1$, $f^{\prime}(x) \rightarrow-\infty, f^{\prime}(2)=\mathrm{e}^{2}-a>0$, hence there exists $x_{0} \in(1,2)$ such that $f^{\prime}\left(x_{0}\right)=0$ $\Rightarrow \mathrm{e}^{x_{0}}=\frac{a}{x_{0}-1}$. At this point, $f(x)$ is monotonically decreasing on $\left(1, x_{0}\right)$ and monotonically increasing on $\left(x_{0},+\infty\right)$,
$f(x)_{\min }=f\left(x_{0}\right)=\mathrm{e}^{x_{0}}-a \ln \left(a x_{0}-a\right)+a$
$=\frac{a}{x_{0}-1}-a \ln \left(a x_{0}-a\right)+a=a\left(\frac{1}{x_{0}-1}+x_{0}+1-2 \ln a\right)>a(4-2 \ln a)>0$, satisfying the condition.
In summary, the range of real number $a$ is $\left(0, \mathrm{e}^{2}\right)$.
10. The function $f(x)=\mathrm{e}^{x}-a \ln (a x-a)+a(a>0)$, if $f(x)>0$ always holds, then the range of the real number $a$ is
$\qquad$.