Solving logarithm problem

Question

If $x=\frac{e^y-e^{-y}}{e^y+e^{-y}}$ show that, $y=\frac{1}{2}\log_{e}\frac{1+x}{1-x}$

What I've tried,

taking log to the base e on both sides, we get;

$\log_{e}{x}$=$\log_{e}{m^2-1}/m^{2}+1$, where $m=e^y$.

But, how should I proceed from here. Any further help or hint is much appreciated.

Answer 1

from $$x=\frac{e^{2y}-1}{e^{2y}+1}$$ we get by multiplication with $e^{2y}+1$ $$e^{2y}+1=\frac{e^{2y}}{x}-\frac{1}{x}$$ from here we get $$e^{2y}\left(1-\frac{1}{x}\right)=-1-\frac{1}{x}$$ for $$x\ne 1$$ we get $$e^{2y}=\frac{x+1}{1-x}$$ can you proceed?

Answer 2

HINt: find $e^{2y}$ from $x=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}$ $$x=\frac{e^y-e^{-y}}{e^y+e^{-y}} =\frac{e^y-e^{-y}}{e^y+e^{-y}}\times \dfrac{e^y}{e^y} \\\to x=\dfrac{e^{2y}-1}{e^{2y}+1}$$then $$xe^{2y}+x=e^{2y}-1 \\e^{2y}(x-1)=-1-x\\\to e^{2y}=\dfrac{1+x}{1-x}$$now apply $\log$

Answer 3

HINT: Clear the denominator, distribute $x$, combine terms with $e^y$ and $e^{-y}$, then you can solve for $e^{2y}$. At this point, you can take natural logarithms.

Answer 4

$$x = \dfrac{e^y - e^{-y}}{e^y + e^{-y}} \\ \implies x = \dfrac{e^{2y}-1}{e^{2y}+1} \quad\text{[Simplify it by multiplying something common both on top and bottom]} \\ \implies e^{2y} = \dfrac{1+x}{1-x} \text{more simplification} \\$$ There you have it

Answer 5

$$\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}=x$$

so that by inversion*

$$e^{2y}=\frac{1+x}{1-x}.$$

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*when $x\ne1$, $$\frac{z-1}{z+1}=x\iff z-1=zx+x\iff z(1-x)=1+x.$$

Answer 6

First of all: we have that $e^y, e^{-y} > 0$, hence $e^{y} - e^{-y} < e^{y} + e^{-y}$, which shows that $x < 1$. Moreover, we have that $(-1)(e^{y} + e^{-y}) = -e^{y} - e^{-y} < e^{y} - e^{-y}$, so we have that $-1 < x$. We conclude that $x \in ]-1,1[$. This piece is needed in order to be able to do all transformations in the following solution.

Suppose $x = \frac{e^y - e^{-y}}{e^y + e^{-y}}$, then we have that $$x(e^y + e^{-y}) = e^y - e^{-y}$$ and grouping the $e^y$ on the right hand side and the $e^{-y}$ on the left side, we find that $$(x - 1)e^y = -(x + 1)e^{-y}.$$ Therefore, we have that (swithching $e^{-y}$ to the right hand side) and the $x -1$ to the left hand side (which is possible since $x \neq 1$): $$e^{2y} = \frac{1 + x}{1 - x}.$$ Taking the natural logarithm of both sides (which we can, since $\frac{1 + x}{1 - x} > 0$ because $x \in ]-1,1[$), we find $$2y = \log_e\frac{1+x}{1 - x}.$$