Inverse of a function with natural log

**BobRoss** · September 22nd 2012, 05:42 PM

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

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.

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

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?

**Prove It** · September 22nd 2012, 05:51 PM

Originally Posted by BobRoss

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

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.

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

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?

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

**BobRoss** · September 22nd 2012, 06:03 PM

What is happening when you go from:

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

to

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

?

**Prove It** · September 22nd 2012, 06:03 PM

Originally Posted by BobRoss

What is happening when you go from:

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

to

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

?

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

**BobRoss** · September 22nd 2012, 06:08 PM

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

**BobRoss** · September 23rd 2012, 12:55 PM

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

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

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

**skeeter** · September 23rd 2012, 01:11 PM

Originally Posted by BobRoss

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

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

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

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

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

solve for $y$

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

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