How to solve this equation: $ x(e^{x}-e^{-x})-e^{x}=0 $ I've already tried to simplify it as:
$$e^{2x}(x-1)-x=0$$
And that gives me: $e^{2x}=\frac{x}{x-1}$
so $2x = \ln(\frac{x}{x-1}) => 2x = \ln(x)-\ln(x-1)$
But then I am stuck. If anyone has a trick or anything?
Thank you in advance.
 A: Divide both sides by $2$ and recall the hyperbolic sine function:
$$x\frac{e^x-e^{-x}}2-\frac{e^x}2=0$$
$$x\sinh(x)-\frac{e^x}2=0$$
Now we employ Newton's method:
$$x_{n+1}=x_n-\frac{x_n\sinh(x_n)-\frac{e^{x_n}}2}{\sinh(x_n)+x_n\cosh(x_n)-\frac{e^{x_n}}2}$$
With $x_0=1$,
$x_1=1.135335283$
$x_2=1.119569316$
$x_3=1.119322977$
$x_4=1.119322917$
And so that is the solution near $x_0=1$.  With $x_0=-0.5$,
$x_1=-0.530772575$
$x_2=-0.530045575$
$x_3=-0.530045160$
$x_4=-0.530045160$
And that is the desired solution near $x=-0.5$
A: with a substitution it looks nicer setting $$t=e^x$$ then we have $$t^2(\ln(t)-1)-\ln(t)=0$$ and now see the Newton-Raphson method
A: Since $e^x-e^{-x}=2\sinh x$ and $e^x=\sinh x+\cosh x$, we can rewrite the equation as $2x\sinh x-\sinh x-\cosh x=0$ or
$$
2x=1+\coth x
$$
(after noticing that $x=0$ is not a solution).
If we consider $f(x)=2x-1-\coth x$, we have
$$
f'(x)=2-(1-\coth^2x)=1+\coth^2x
$$
which is positive. So the function is increasing in both the intervals $(-\infty,0)$ and $(0,\infty)$.
Since
$$
\lim_{x\to-\infty}f(x)=-\infty,\quad
\lim_{x\to0^-}f(x)=\infty,\quad
\lim_{x\to0^+}f(x)=-\infty,\quad
\lim_{x\to\infty}f(x)=\infty
$$
we know that the equation has two solutions.
