# Functions + Natural Logs

1. Dec 2, 2007

1. The problem statement, all variables and given/known data
The function f is defined for $$x>2$$ by $$f(x)=\ln x+\ln(x-2)-\ln(x^{2}-4)$$
a. Express f(x) in the form of $$(\ln\frac{x}{x+a})$$

b. Find an expression for $$f^{-1}(x)$$
2. Relevant equations
..

3. The attempt at a solution

Well, I simplified it to:

$$f(x)=\ln(\frac{x^{2}-2x}{x^{2}-4})$$

I can't figure what to do, ill keep thinking

Last edited: Dec 2, 2007
2. Dec 2, 2007

### Dick

I hope you mean you simplified it to log((x-2)/(x^2-4)). Factor the denominator. What was for dinner?

3. Dec 2, 2007

Im sorry, I left an ln x out, for some reason I typed \lnx and it appeared as nothing. Should I still factor the denominator? ~ I had chow mein with chicken :D

4. Dec 2, 2007

### Dick

I love chicken chow mein. You can still factor the denominator and cancel the x-2 in the numerator.

5. Dec 2, 2007

Oh I get it!

So $$f(x)=\ln \frac {x(x-2)}{(x+2)(x-2)}$$

So $$f(x)=\ln \frac{x}{x+2}$$

Awesome, part A solved. Now find the inverse..
Give me a sec here

Argh, the only way i know of obtaining inverses is by flipping x and y's.. here it seems a tidbit different. I know that:

The domain of the inverse is the range of the original function.
The range of the inverse is the domain of the original function.

How am I supposed to find the inverse :S, I get that $$f^{-1}(x)=(e^{x})(y+2)..$$

Last edited: Dec 2, 2007
6. Dec 2, 2007

Any hints?

7. Dec 3, 2007

### HallsofIvy

Staff Emeritus
No, a function of "x" cannot have a "y" in it!
You have to start with $$y= ln(\frac{x}{x+2})$$
The first step is, just as you say, to "swap" x and y:
$$x= ln(\frac{y}{y+2}$$
Now solve for y. You need to get rid of the log:
$$e^x= \frac{y}{y+2}$$
Now get rid of the fraction by multiplying both sides by y+ 2.
$$e^x(y+2)= y$$
This is where you stopped, right? But that "y" on the right is NOT the inverse function because you still haven't solved for y.
$$e^xy + 2e^x= y$$
Now get y by itself on the left, with no y on the right. Can you do that?
(How would you solve ay+ b= y for y?)