# The Inverse of ln

1. Feb 24, 2009

### hostergaard

1. The problem statement, all variables and given/known data
http://img5.imageshack.us/img5/2327/nummer1.jpg [Broken]

2. Relevant equations

5a: ln((x*(x-2))/(x2-4))
ln((x22x)/(x2-4))
ln(x/(x+2)) <-- the answer (i think)
3. The attempt at a solution
Im not completly sure about that answer in 5a so could somebody check that? also I'am having truble with b:

f(x)--> y=ln(x/(x+2))
f-1(x)--> x=ln(y/(y+2))
ex=y/(y+2)

then i get in truble...

Last edited by a moderator: May 4, 2017
2. Feb 24, 2009

### HallsofIvy

Staff Emeritus
$x^2- 4= (x-2)(x+2)$ and, as long as x is not equal to 2, you can cancel the two (x- 2) factors. But the original formula is not defined for x= 2 so, yes, that is equivalent to the original.

Good, you are almost done. If you had A= y/(y+ 2) you could solve for y by multiplying on both sides by y+ 2 to get Ay+ 2A= y so (1- A)y= 2A and y= 2A/(1- A). Does that help?

Last edited by a moderator: May 4, 2017
3. Feb 24, 2009

ohh, i see!