= Solution
Fix $\varepsilon>0$. Since a rate-one exponential variable has
$\mathbb P(X_n>x)=e^{-x}$,
$$
\sum_n\mathbb P\{X_n>(1+\varepsilon)\log n\}
=\sum_n n^{-(1+\varepsilon)}<\infty.
$$
The first Borel--Cantelli lemma gives the upper bound
$\limsup X_n/\log n\leq1$. Conversely,
$$
\sum_n\mathbb P\{X_n>(1-\varepsilon)\log n\}
=\sum_n n^{-(1-\varepsilon)}=\infty.
$$
These events are independent, so the second lemma makes them occur infinitely often. Hence the lower bound is $1-\varepsilon$. Letting rational $\varepsilon\downarrow0$ proves
$$
\boxed{\limsup_{n\to\infty}\frac{X_n}{\log n}=1}
\quad\hbox{almost surely}.
$$
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