Solve Inverse of ln: x=ln(y/(y+2))

[hostergaard]

Homework Statement

http://img5.imageshack.us/img5/2327/nummer1.jpg

Homework Equations

5a: ln((x*(x-2))/(x²-4))
ln((x²2x)/(x²-4))
ln(x/(x+2)) <-- the answer (i think)

The Attempt at a Solution

Im not completely 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))
e^x=y/(y+2)

then i get in truble...

 

[HallsofIvy]

Homework Statement

http://img5.imageshack.us/img5/2327/nummer1.jpg

Homework Equations

5a: ln((x*(x-2))/(x²-4))
ln((x²2x)/(x²-4))
ln(x/(x+2)) <-- the answer (i think)

[itex]x^2- 4= (x-2)(x+2)[/itex] 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.

The Attempt at a Solution

Im not completely 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))
e^x=y/(y+2)

then i get in truble...

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?

 

[hostergaard]

ohh, i see!
Then the answer is 2e^x/(1-e^x)
You sir, are very helpful (wo?)man.
Thanks a lot!