# Thread: Solving Y in terms of X

1. ## Solving Y in terms of X

i'm having some trouble remembering how to solve this in terms of x.
$\displaystyle y=(e^x-e^-^x)/2$

This is what i was thinking, but, i'm not sure if it is right or not.
$\displaystyle x=(e^y-e^-^x)/2$
$\displaystyle 2x=(e^y-e^y)$
$\displaystyle log(2x)=2y*log(e)$
$\displaystyle log(2x)/(2*log(e))=y$

$\displaystyle y=log(2x)/(2*log(e))$
I feel like this is not the right answer though :/

2. Originally Posted by fenderic
i'm having some trouble remembering how to solve this in terms of x.
$\displaystyle y=(e^x-e^-^x)/2$

This is what i was thinking, but, i'm not sure if it is right or not.
$\displaystyle x=(e^y-e^-^x)/2$
$\displaystyle 2x=(e^y-e^y)$
$\displaystyle log(2x)=2y*log(e)$
$\displaystyle log(2x)/(2*log(e))=y$

$\displaystyle y=log(2x)/(2*log(e))$
I feel like this is not the right answer though :/
I think the answer you're looking for is y = sinh(x) (or is it cosh(x), it's been awhile since I studied calculus).

3. Hello, fenderic!

i'm having some trouble remembering how to solve this in terms of x.

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

It is already "solved in terms of $\displaystyle \,x."$
I must assume you mean: "Solve for $\displaystyle \,x$ in terms of $\displaystyle \,y."$

We have: .$\displaystyle \dfrac{e^x - e^{-x}}{2} \;=\;y \quad\Rightarrow\quad e^x - e^{-x} \;=\;2y$

Multiply by $\displaystyle e^x\!:\;\;e^{2x} - 1 \;=\;2ye^x \quad\Rightarrow\quad (e^x)^2 - 2y(e^x) - 1 \;=\;0$

Quadratic Formula: .$\displaystyle e^x \:=\:\dfrac{2y \pm\sqrt{4y^2+4}}{2} \quad\Rightarrow\quad e^x \;=\;y \pm \sqrt{y^2+1}$

Since $\displaystyle e^x$ is always positive, we take the positive root: .$\displaystyle e^x \;=\;y + \sqrt{y^2+1}$

,. . Therefore: .$\displaystyle x \;=\;\ln\left(y + \sqrt{y^2+1}\right)$

4. thank you Soroban!