= Solution
The chord through the endpoints is the <straight line> $y=(x-1)\log2$. At $\xi=1/\log2$, the <tangent line> is
$$
y=\log\xi+\frac{x-\xi}{\xi}
=\log\xi+(x-\xi)\log2.
$$
These two <straight lines> have the same <slope>, as required by the <mean value theorem>. Since $f''(x)=-1/x^2<0$, the <natural logarithm> is a <concave function>: its graph lies above the chord and below this <tangent line>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2-q3-mean-value.png]
{title=The mean value theorem for the logarithm on the interval from one to two}
{description=The chord and the <tangent line> at $\xi=1/\log2$ are parallel. The marked point lies on the graph of the <natural logarithm>.}
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