Finding the Inverse of a Function

[matadorqk]

Homework Statement

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)

Homework Equations

..

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

 

[Dick]

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

 

[matadorqk]

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

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

 

[Dick]

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

 

[matadorqk]

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

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)..

 

[matadorqk]

Any hints?

 

[HallsofIvy]

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)..

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?)