Thread: Inverse of equation involving e^(-x)

1. Inverse of equation involving e^(-x)

Hi,

My problem is to find the inverse of $\displaystyle y = f(x) = \frac{1}{2}(e^x - \frac{1}{e^x})$.

I can get up to using $\displaystyle v$ for $\displaystyle e^x$, and that $\displaystyle v > 0$

$\displaystyle 2yv = v^2 -1$

$\displaystyle 0 = v^2 - 2yv - 1$

The using the quadratic formula I get $\displaystyle v = \frac{2y \pm \sqrt{(-2y)^2 + 4}}{2}.$

Here I am stuck

2. Following on from where you left off...

$\displaystyle v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $\displaystyle (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$)

Note that $\displaystyle 2 = \sqrt{4}$ so we get

$\displaystyle v = y \pm \sqrt{(y^2 + 1)}$

Now since $\displaystyle v = e^x$, take natural logs of both sides to get...

$\displaystyle x = \ln(v) =$ ..? Can you finish?

3. Yes.

I kept going around in circles about what to do with the square root in the quadratic formula.

All is good now. Thanks

Following on from where you left off...

$\displaystyle v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $\displaystyle (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$)

Note that $\displaystyle 2 = \sqrt{4}$ so we get

$\displaystyle v = y \pm \sqrt{(y^2 + 1)}$

Now since $\displaystyle v = e^x$, take natural logs of both sides to get...

$\displaystyle x = \ln(v) =$ ..? Can you finish?
I'll add a small note: One of the solutions (either the positive or the negative root) must be discarded ....