Take the function $\ln(3 - x)$. By the logic of transformations that I have been taught, the order of transformations goes: $\ln(x)$ to $\ln(-x)$ which reflects the curve across the $y$-axis, then $\ln(-x + 3)$. This additional $+3$ should then push the curve to the left, hence turn the asymptote to $x = -3$. (This is what I show in the black curve.) However, the way the graph is shown in my book is that the transformation actually makes the asymptote $x = 3$, which disagrees with me, although the direction of the graph relative to then $x$-axis stays is the same as mine.
Why? Someone please explain. I asked my teacher, and he said 'yea that's weird', and I am yet to find an answer.
Elaborating on @Dave's comment, here's the sketch of what is going on.
You know that if you have a function $f(x)$, then $f(-x)$ reflects the function over the $y$-axis.
You know that if you have a function $f(x)$, then $f(x+3)$ shifts the function by $3$ to the left.
Now, consider your function:
Starting with $f(x)=\ln(x)$, you look at $f(-x)=\ln(-x)$, which reflects the function over the $y$-axis.
Now, let's call the new function we're dealing with $g$. In other words, $g(x)=\ln(-x)$. We would like to shift $3$ to the left, so we look at $g(x+3)=\ln(-(x+3))=\ln(-x-3)$.
If you want to shift to the right by $3$, we could apply the change $g(x-3)=\ln(-(x-3))=\ln(3-x)$, which is what you get above.
The trick is that when you substitute $x+3$, you can't put the "$+3$" in anywhere, you have to replace $x$ by $x+3$ wherever it appears and remember to distribute. Therefore, in your case, the negative in front of the $x$ must be distributed to the $+3$ as well.