If $\displaystyle x=\frac{e^t-e^{-t}}{2}$, solve for t.
This is my work.
$\displaystyle 2x=e^t-e^{-t}$
$\displaystyle \ln 2x=\ln (e^t-e^{-t})$
I'm stuck trying to simplify $\displaystyle \ln (e^t-e^{-t})$. Help?
Thanks.
If $\displaystyle x=\frac{e^t-e^{-t}}{2}$, solve for t.
This is my work.
$\displaystyle 2x=e^t-e^{-t}$
$\displaystyle \ln 2x=\ln (e^t-e^{-t})$
I'm stuck trying to simplify $\displaystyle \ln (e^t-e^{-t})$. Help?
Thanks.
$\displaystyle x = \frac{e^t - e^{-t}}{2}
$
$\displaystyle 2x = e^t - e^{-t}
$
$\displaystyle 2xe^t = e^{2t} - 1$
$\displaystyle 0 = e^{2t} - 2xe^t - 1$
$\displaystyle e^t = \frac{2x \pm \sqrt{4x^2 + 4}}{2}
$
$\displaystyle e^t = \frac{2x \pm 2\sqrt{x^2 + 1}}{2}$
$\displaystyle e^t = x \pm \sqrt{x^2+1}$
since $\displaystyle \sqrt{x^2+1} > x$ ...
$\displaystyle e^t = x + \sqrt{x^2+1}
$
$\displaystyle t = \ln\left(x + \sqrt{x^2+1}\right)$
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 $\displaystyle e^{-t}= \frac{1}{e^t}= \frac{1}{y}$, your equation becomes $\displaystyle x= \frac{y+ 1/y}{2}$. Multiply both sides by 2y to get [tex]2xy= 2y^2+ 1[tex] or $\displaystyle 2y^2- 2xy+ 1= 0$. Use the quadratic formula to solve that for y and then take t= ln(y).