Question

Practice with Logarithmic Inverse functions! For fun :)

\[f(x) = \frac{e^{2x}-1}{2-e^{2x}}\]

Answer 1

Lmao.

Answer 2

1/2 ln something..

give me a sec xD

Answer 3

1/2 ln ((2x+1)/(x+1)) ?

Answer 4

Well, you dont even need to switch the variables first before simplifying I guess.

Answer 5

i'm used to it.
teachers and their solutions.
need to do their way if you want the mark :P

Answer 6

\[y=\frac{e^{2x}-1}{2-e^{2x}}\]\[2y-ye^{2x} = e^{2x}-1\]\[2y+1=e^{2x}+ye^{2x}\]\[e^{2x}(y+1)=2y+1\]\[e^{2x}=\frac{2y+1}{y+1}\]\[\ln e^{2x} =\ln\left(\frac{2y+1}{y+1}\right)\]\[2x=\ln\left(\frac{2y+1}{y+1}\right)\]\[f^{-1}(x) = \ln\left[\left(\frac{2x+1}{x+1}\right)^{1/2} \right]\]