Sketching the graph of $\ln(4-x)$ by transforming $\ln(x)$

Question

Using the base graph of $\ln(x)$: I "reflect" the graph of $\ln(x)$ in the $y$-axis, giving me $\ln(-x)$, the graph now cuts the $x$-axis at $(-1,0)$.

I then translate the graph by negative $4$ units along the $x$-axis, so the graph now cuts the $x$-axis at $(-5,0)$.

This is wrong, could anyone tell me why that is? Thanks!

Answer 1

The zero point of $x-4$ is at $x=4$, so when you substitute $4-x$ for $x-4$, you’re reflecting in the line $x=4$. This has nothing to do with the logarithm function: look, for instance, at $f(x)=4-x$ instead. If you reflect $y=x$ in the $y$-axis and shift left $4$ units, you end up with the graph of $y=-x-4$, not $y=4-x$: you’ve reflected $y=x+4$ in the $y$-axis instead of $y=x-4$.

To put it a little differently, when you replace $-x$ by $4-x$, you’re not shifting the $x$-coordinate $4$ units to the left: you’re shifting the $(-x)$-coordinate $4$ units to the left, which shifts the $x$-coordinate $4$ units to the right. And that puts your $x$-intercept where it belongs, at $3$.

Answer 2

Because the $x$ is negative after your reflection, you need to translate $+4$ units along the $x$ axis, making it cut at $(3,0)$