Why do logarithmic curves not follow the general rules for transformations?

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.

• $\ln(-x+3)=\ln(-(x-3))$
– Dave
May 21, 2017 at 14:41
• This basically implies that you go from ln(x) to ln(x-3), which is shifting the curve to the right by 3, and then we reflect that in the y-axis. Still gives the black curve in my picture, and not the correct red curve. Please point out which part of my logic is wrong. This is how I've been taught. May 21, 2017 at 14:48
• The process you just described does not result in the black curve, but rather the red one. If you transform $\ln(x)$ to $\ln(x-3)$, the curve indeed shifts to the right $3$ units. However, when you know transform $\ln(x-3)$ to $\ln(-(x-3))$, the curve simply reflects across the line $x=3$ (since this line acts like the new $y$ axis for $\ln(x-3)$), and does not create a new asymptote at $x=-3$ (i.e. the reflection of $\ln(x-3)$ to $\ln(-(x-3))$ does not reflect across the line $x=0$).
– Dave
May 21, 2017 at 14:53
• That's weird, I know I'd be taking up your time, but may I please ask that you explain why the line x=3 becomes the 'new y-axis'? May 21, 2017 at 14:56
• What I mean by "like the new $y$ axis" is just that the line $x=3$ is the reflection line for $\ln(x-3)$ to $\ln(-(x-3))$, much like how the $y$ axis (i.e. $x=0$) is the reflection line for $\ln(x)$ to $\ln(-x)$.
– Dave
May 21, 2017 at 14:58

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.

• 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.
@Dave more or less sums it up in the comment above. When you're changing $-x$ into $3-x$, you can also see that as changing $x$ into $x-3$. This should and does move the graph to the right.