Once you have $g(x)$ written as a single $\ln$ of some expression, think about how changes in that inside expression affect the value of $\ln$.

After combining the logs, you should get $\ln$ of a product like $(x+2)(6-x)$. Expand this product to get a quadratic, and then use what you know about parabolas and their vertices to find where this quadratic is largest.

For a downward-opening parabola $ax^2+bx+c$, the maximum occurs at $x = -\dfrac{b}{2a}$. Apply this to your quadratic, and make sure the $x$-value you find is between $-2$ and $6$.

Type g(x) = ln(x+2) + ln(6-x) into Desmos. Make sure the ln function is used and that the expression is exactly as given.

Look at the graph; you should see it only between $x=-2$ and $x=6$ (the graph will not appear outside this domain because the logs are undefined there).

Click or tap on the highest point of the graph, or use Desmos’s maximum feature (click the graph, then choose maximum if available). Read off the $x$-coordinate of that highest point; that $x$-value is where $g$ attains its maximum.

Use the log rule $\ln a + \ln b = \ln(ab)$ to rewrite the function:

So $g(x)$ is the natural log of the product $(x+2)(6-x)$.

The natural log function is increasing on its domain: if $u>v>0$, then $\ln u > \ln v$.

That means $g(x)$ is largest exactly when the inside of the log, $(x+2)(6-x)$, is largest (for $-2<x<6$).

So we only need to maximize the function

First expand $h(x)$:

This is a quadratic with $a=-1$, $b=4$, and $c=12$. Because $a<0$, its graph is a downward-opening parabola, so it has a maximum at its vertex.

The $x$-coordinate of the vertex of a quadratic $ax^2+bx+c$ is given by

We will now apply this formula to $h(x)$.

For $h(x) = -x^2 + 4x + 12$, we have $a=-1$ and $b=4$. Plug into the vertex formula:

So $h(x)$ is maximized at $x=2$, which means $g(x)$ also attains its maximum value at

$x = 2$.