6.4. (New York, 80). Find all pairs of numbers \(a > 1\), \(b > 0\), for which the equation \(a^x = x^b\) has exactly one positive solution. Indicate this solution for each pair of found values \(a, b\).
6.4. The equation $a^{x}=x^{b}$ is equivalent to the equation $c^{x}=x$, where $c=a^{1 / b}>1$. Since the derivative of the function $f(x)=c^{x}-x$, which is $f^{\prime}(x)=c^{x} \ln c-1$, is positive when $c^{x}>\log _{c} e$ and negative when $c^{x}<\log _{c} e$, there exists a unique $x_{0}>0$ that satisfies the equation $c^{x_{0}}=\log _{c} e$. If $f\left(x_{0}\right)>0$, then the original equation has no solutions; if $f\left(x_{0}\right)=0$, then the equation has a unique positive solution $x=x_{0}$; if $f\left(x_{0}\right)<0$, then the equation has two positive solutions (since $f(0)=-1<0$ and $\left.\lim _{x \rightarrow \infty} f(x)=\infty\right)$.

Thus, the original equation meets the problem's requirements if and only if $x_{0}=c^{x_{0}}=\log _{c} e_{\mathrm{1}}$, i.e., $x_{0}=c^{\log _{c} e}=e$ and $c=e^{1 / e}$.

Therefore, the sought values of $a, b$ (and only they) satisfy the equation $a^{e}=e^{b}$, where $b>0$, i.e., $a=t, b=e \ln t$, where $t \in(1 ; \infty)$, and the unique positive solution of the equation is $x=$.