logarithmic confusion

Question

I need to find the inverse function of the below equation.

$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

When I apply ln to both sides I get the wrong answer. I get the result of ylne - (-y)lne - (ylne-ylne) = 2y

lnx = 2y .

Why does this not work?

The way the solution is shown is by starting multiplying e^x to both the numerator and denominator which ends up giving a completely different answer.

Answer 1

It looks like you're doing this:

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

but $\ln$ doesn't distribute across addition, so that would be incorrect.

Answer 2

You could do it with hyperbolic functions. See Wikipedia.

$\sinh x = \frac{e^x - e^{-x}}{2}$

$\cosh x = \frac{e^x + e^{-x}}{2}$

So, your expression is $\tanh x$ and its inverse is $\tanh^{-1} x$.

Answer 3

Hint:

$y=f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}= \sinh (x) / \cosh(x) = \tanh(x)$

and $x = arctanh(y) =\frac{1}{2} \ln ((1+y)/(1-y))$

re. to inverse hyperbolic f.

Answer 4

The idea of $\log(a + b) = \log (a) + \log(1 + \frac{b}{a})$ is to isolate $a$ from $b$. As you can see $a$ was somewhat isolated as $\log (a)$, but $a$ & $b$ are still coupled as a ratio. I thought this transform would be useful for approximation (bounding), summation, asymptotic analysis.

I came up with a similar transform with trigonometric functions, which some people thought was useful in Polar Coordinates. I don't know as I don't work in it.

Answer 5

$$y=\frac{e^{2x}-1}{e^{2x}+1}$$ $$\frac{y+1}{y-1}=\frac{e^{2x}}{-1}$$ Replace x with y and y with x. $$e^{2y}=\frac{1+x}{1-x}$$ $$2y=ln(\frac{1+x}{1-x})$$ $$f^{-1}(x)=ln(\frac{1+x}{1-x})$$