# Thread: Finding the inverse of a function

1. ## Finding the inverse of a function

Hi. For my homework one of the first things I'm supposed to be able to do is find the inverse of

y = (4+e^x)/(4-e^(-x))

I don't get how this is possible. When I try to find the natural log of both sides I end up with

lny = ln(4+e^x) - ln(4-e^x)

There must be some algebra I'm not getting here, because I know there's no log rule that can help me get that X out of the ln

Thanks, the homework is due tomorrow

2. Originally Posted by chubigans
Hi. For my homework one of the first things I'm supposed to be able to do is find the inverse of

y = (4+e^x)/(4-e^(-x))

I don't get how this is possible. When I try to find the natural log of both sides I end up with

lny = ln(4+e^x) - ln(4-e^x)

There must be some algebra I'm not getting here, because I know there's no log rule that can help me get that X out of the ln

Thanks, the homework is due tomorrow
Solve the equation $y=\dfrac{4+e^x}{4-e^{-x}}$ for x. Afterwards swap the variables:

$y=\dfrac{4+e^x}{4-e^{-x}}~\implies~4y-ye^{-x}=4+e^x$ Multiply through by $e^x$

$4ye^x-y=4e^x+e^{2x}~\implies~e^{2x} +4e^x(1-y)+y=0$

This is a quadratic equation in $e^x$:

$e^x=-2(1-y)\pm\sqrt{4(1-y)^2-y}$

Therefore:

$x=\ln\left( 2y-2 + \sqrt{4y^2-9y+4} \right)~\vee~x=\ln\left( 2y-2 - \sqrt{4y^2-9y+4} \right)$

Now swap the variables and you'll get the inverse of the function.

3. Just wanted to say thank you, I can't believe I didn't see that!