# Thread: How do I solve for x?

1. ## How do I solve for x?

If $x=\frac{e^t-e^{-t}}{2}$, solve for t.

This is my work.

$2x=e^t-e^{-t}$
$\ln 2x=\ln (e^t-e^{-t})$

I'm stuck trying to simplify $\ln (e^t-e^{-t})$. Help?

Thanks.

2. Originally Posted by chengbin
If $x=\frac{e^t-e^{-t}}{2}$, solve for t.

This is my work.

$2x=e^t-e^{-t}$
$\ln 2x=\ln (e^t-e^{-t})$

I'm stuck trying to simplify $\ln (e^t-e^{-t})$. Help?

Thanks.
$x = \frac{e^t - e^{-t}}{2}
$

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

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

$0 = e^{2t} - 2xe^t - 1$

$e^t = \frac{2x \pm \sqrt{4x^2 + 4}}{2}
$

$e^t = \frac{2x \pm 2\sqrt{x^2 + 1}}{2}$

$e^t = x \pm \sqrt{x^2+1}$

since $\sqrt{x^2+1} > x$ ...

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

$t = \ln\left(x + \sqrt{x^2+1}\right)$

3. Thanks for that solution.

Is there a way to move on like the way I did it?

4. No, your method was making things more complicated rather than simplifying. One thing that might make the solution easier to understand is to let [tex]y= e^t[tex] to start with. Since $e^{-t}= \frac{1}{e^t}= \frac{1}{y}$, your equation becomes $x= \frac{y+ 1/y}{2}$. Multiply both sides by 2y to get [tex]2xy= 2y^2+ 1[tex] or $2y^2- 2xy+ 1= 0$. Use the quadratic formula to solve that for y and then take t= ln(y).