= Solution
For $f=x\ln x$, the <derivatives> are $f'=1+\ln x$ and $f''=1/x>0$. Its unique <stationary point> is the global <minimum> $(e^{-1},-e^{-1})$. For $g=\ln x/x$, $g'=(1-\ln x)/x^2$ changes from positive to negative at $x=e$, giving the global <maximum> $(e,e^{-1})$. Their endpoint <limits> are
$$
\boxed{\begin{array}{c|cc}&x\to0^+&x\to\infty\\\hline f(x)&0\text{ from below}&+\infty\\g(x)&-\infty&0\text{ from above}\end{array}}
$$
An intersection requires $(x^2-1)\ln x=0$, and on $x>0$ both factors vanish only at $x=1$. There $f=g=0$ and $f'=g'=1$, proving a common <tangent line> $y=x-1$. In fact,
$$
f-g=\frac{(x^2-1)\ln x}{x}>0\quad(x>0,\ x\ne1),
$$
because the two numerator factors always have the same sign. Thus the curves touch at $(1,0)$ without exchanging their order.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1-logarithmic-tangency.png]
{title=Two logarithmic curves touching at their unique intersection}
{description=Original common-axis plot marking the extrema and the shared tangent at $(1,0)$. The curve $x\ln x$ lies above $\ln x/x$ everywhere else on the positive axis.}
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