7. 10. Given the function $f(x)=-\frac{\ln x}{x}+e^{a x-1}$ has a minimum value of $a$, then the minimum value of $a$ that satisfies the condition is $\qquad$
7. $-\frac{1}{e^{2}}$

Analysis: Clearly $x>0$, and we are familiar with the inequality $e^{x} \geq x+1$ (equality holds if and only if $x=0$);
Therefore, $f(x)=\frac{x e^{a x-1}-\ln x}{x}=\frac{e^{a x+\ln x-1}-\ln x}{x} \geq \frac{(a x+\ln x-1)+1-\ln x}{x}=a$
Equality holds if and only if $a x+\ln x-1=0$, i.e., $a=\frac{1-\ln x}{x}$, where $x>0$. Let $g(x)=\frac{1-\ln x}{x}$, and we will find the minimum value of $g(x)$. $g^{\prime}(x)=\frac{\ln x-2}{x^{2}}=0$, yielding $x=e^{2}$, which is easily identified as a point of minimum (also the minimum value point), and $g\left(e^{2}\right)=-\frac{1}{e^{2}}$, hence the minimum value of $a$ is $-\frac{1}{e^{2}}$.