Inverse of a function with natural log

I am asked to find the inverse of $\displaystyle f(x)=ln(x+\sqrt{x^2 +1})$

I understand the basics of how to find the inverse but I get stuck on solving this one. Here is what I have so far.

$\displaystyle y=ln(x+\sqrt{x^2 +1})$

$\displaystyle e^y =x+ \sqrt{x^2+1}$

I'm not sure how to proceed from here. I thought maybe squaring both sides would be the next step but I can never come up with the correct answer. How do I solve this?

Re: Inverse of a function with natural log

Quote:

Originally Posted by

**BobRoss** I am asked to find the inverse of $\displaystyle f(x)=ln(x+\sqrt{x^2 +1})$

I understand the basics of how to find the inverse but I get stuck on solving this one. Here is what I have so far.

$\displaystyle y=ln(x+\sqrt{x^2 +1})$

$\displaystyle e^y =x+ \sqrt{x^2+1}$

I'm not sure how to proceed from here. I thought maybe squaring both sides would be the next step but I can never come up with the correct answer. How do I solve this?

$\displaystyle \displaystyle \begin{align*} e^y &= x + \sqrt{x^2 + 1} \\ e^y - x &= \sqrt{x^2 + 1} \\ \left(e^y - x \right)^2 &= x^2 + 1 \\ e^{2y} - 2x\,e^y + x^2 &= x^2 + 1 \\ e^{2y} - 2x\,e^y &= 1 \\ e^{2y} - 1 &= 2x\,e^y \\ e^y - e^{-y} &= 2x \\ \frac{1}{2}\left( e^y - e^{-y} \right) &= x \end{align*}$

Re: Inverse of a function with natural log

What is happening when you go from:

$\displaystyle e^{2y} -1=2xe^y$

to

$\displaystyle e^y - e^{-y} =2x$

?

Re: Inverse of a function with natural log

Quote:

Originally Posted by

**BobRoss** What is happening when you go from:

$\displaystyle e^{2y} -1=2xe^y$

to

$\displaystyle e^y - e^{-y} =2x$

?

Dividing both sides by $\displaystyle \displaystyle \begin{align*} e^y \end{align*}$.

Re: Inverse of a function with natural log

Ahh right that's what I thought but I just got confused for a second. Thanks, this helps a ton!

Re: Inverse of a function with natural log

I have another inverse function question that I am having trouble with.

$\displaystyle f(x)=arctan(\frac{x-1}{x+1})$

I don't even really know how to start solving this one.

Re: Inverse of a function with natural log

Quote:

Originally Posted by

**BobRoss** I have another inverse function question that I am having trouble with.

$\displaystyle f(x)=arctan(\frac{x-1}{x+1})$

I don't even really know how to start solving this one.

$\displaystyle x = \arctan\left(\frac{y-1}{y+1}\right)$

$\displaystyle \tan{x} = \frac{y-1}{y+1}$

solve for $\displaystyle y$

... in future, start a new problem with a new thread.