Solving $\sinh x = kx$ Can we solve the equation $\sinh x = kx$ for $x$ in terms of elementary functions? I've tried reexpressing the hyperbolic sine as exponentials and converting the equation into a quadratic in $e^x$, but this doesn't seem to make the problem any easier. I've considered expanding $\sinh x$ as a Taylor series, but this doesn't seem useful either.
 A: As mentioned in the comments, the only solution for $k\le1$ is $x=0$ and for $k>1$ there are 3 solutions: $x=0,\pm x_\star$. Although there is no closed form in terms of special functions such as the Lambert W function known, it is not hard to numerically compute $x_\star$, the positive nonzero solution. For example, we have fixed-point iteration:
$$x_{n+1}=\ln(2kx_n+\exp(-x_n))$$
or Newton's method:
$$x_{n+1}=x_n-\frac{\sinh(x_n)-kx_n}{\cosh(x_n)-k}$$
or any other numerical method.
A: As said in comments and answers, you will need a numerical method.
In any manner, instead of looking for the zero of
$$f(x)=\frac{\sinh (x)}{x}-k$$ it would be better to look for the zero of
$$g(x)=\log \left(\frac{\sinh (x)}{x}\right)-\log(k)$$ which is much more linear when $x$ is large.
Newton iterates would be
$$x_{n+1}=x_n-\frac{x_n \sinh (x_n) \log \left(\frac{\sinh (x_n)}{x_n}\right)}{x_n \cosh (x_n)-\sinh (x_n)}$$
If $k$ is large and $x$ too, we could approximate
$$g(x)\sim\log \left(\frac{e^x}{2x}\right)-\log(k)\implies x\sim -W_{-1}\left(-\frac{1}{2 k}\right)$$ where appears the second branch of Lambert function (which is not elementary but simple to use).
Trying for $k=123.456789$, starting with the above estimate as $x_0$, Newton iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 7.5276168997015533341 \\
 1 & 7.5276172335102603321 \\
 2 & 7.5276172335102591983
\end{array}
\right)$$
When $k$ is small, a very good approximation of $g(x)$ could be obtained using the $[4,4]$ Padé approximant of it. This will be
$$g(x)\sim \frac{21 x^2 \left(31 x^2+570\right)}{58 x^4+6300 x^2+71820}- \log(k)$$ which leads to a quadratic equation in $x^2$.
For $k=3$, the above gives $x=2.83971$ while the exact solution would be $x=2.83845$.
