Natural log graph

The graph of ln(4-x); I've thought of this in two ways; ln(-x+4) and ln -(x-4), using the former doesn't it reflect the graph in the y axis and then shift the graph 4 units in the negative x direction...

Let's try that.

Start with $f(x) = \ln x$

"Reflect the graph in the y axis" - this is a transformation $f(x)\rightarrow f(-x)$

$f(-x) = \ln (-x)$

Call this new function $g(x)$ then "shift the graph 4 units in the negative x direction" is a transformation $g(x)\rightarrow g(x+4)$.

$g(x) = \ln (-x)$

$g(x+4) = \ln -(x+4) = \ln (-x -4)$

Can you see your mistake?

In the answers, it shows the graph crossing at x=3, but ln(-x-4) would cross at x=5 so I'm still confused.

The question asks you about ln(4-x) not ln(-x-4). My last post was to show you that if you apply the two transformations that you thought were correct to ln(x), you end up with ln(-x-4) and not ln(4-x). So your reasoning was incorrect.

Does that make sense?

Yes, but how would I correctly apply transformations for this question?

You need to forget about the ln(-x+4) and instead pull out the minus sign as you had already done: i.e. ln[-(x-4)]

Now you can apply the appropriate transformations: Reflexion in the y-axis and translation of 4 units in the x direction.

Is there an order for transformation of graphs?

For a function of the form f(-(x+c)) where c is a constant, you should reflect first then translate.

In your scenario, you have f(-(x-4)) with f(x) = lnx.