Once you have $f(x)$ as $\ln$ of one expression, ask: to make $\ln$ as large as possible, what needs to be true about the expression inside the logarithm?

Look at $(x-1)(7-x)$: where is this product largest for $1<x<7$? Consider its zeros at $x=1$ and $x=7$ and the symmetry of a parabola between these points.

For a downward-opening quadratic with zeros at 1 and 7, where along the interval from 1 to 7 would you expect the highest point of the graph to be?

In Desmos, type f(x) = ln(x-1) + ln(7-x) to graph the function. You should see a curve defined only between $x=1$ and $x=7$.

Zoom or pan so you can clearly see the top of the curve between $x=1$ and $x=7$. Click on the graph near the top; Desmos will show a point labeled with the local maximum. Note the x-coordinate of this maximum point.

You can also graph g(x) = (x-1)(7-x) in Desmos. Use the maximum feature or click the top of this parabola. The x-coordinate of this maximum should match the x-coordinate of the maximum of $f(x)$.

Use the log rule $\ln a + \ln b = \ln(ab)$ for positive $a$ and $b$.

So

The domain $1<x<7$ ensures $x-1>0$ and $7-x>0$, so the logs are defined.

The natural log function $\ln u$ is increasing for $u>0$: if $u_1>u_2>0$, then $\ln(u_1) > \ln(u_2)$.

That means:

• $f(x)$ is largest exactly when the inside of the log, $(x-1)(7-x)$, is largest.

So we just need to maximize

for $1<x<7$.

Rewrite the product a bit:

• $g(x) = (x-1)(7-x)$ is a quadratic function.
• Its value is $0$ when $x=1$ or $x=7$, because one factor becomes $0$.

A quadratic with two real zeros and a negative leading coefficient (you can expand to see it is $-x^2+8x-7$) is a downward-opening parabola:

• It is symmetric about the midpoint of its zeros.
• Its maximum occurs at that midpoint between $x=1$ and $x=7$.

The maximum of $g(x)$, and therefore of $f(x)$, occurs at the midpoint between the zeros $x=1$ and $x=7$.

Compute the midpoint:

So $f(x)$ attains its maximum value when $x=4$, which corresponds to choice C) $4$.