how to solve $\log{x}=cx^4$ for $x$

Question

I was wondering if there is a general solution for this form of equations: $$\log{x}=cx^4$$

Tried: $$ x = e^{cx^4}\\ xe^{-cx^4}=1$$

Answer 1

Note that $\log x/x^4$ has derivative $\frac{1-4\log x}{x^5}$, so the maximum real value for $c$ is at $x=e^{1/4}$, which is a maximum value of $c=\frac{1}{4e}$.

This is one of those equations that requires the Lambert W-function.

Raise your last equation to the fourth power, then multiply by $-4c$ and you get:

$$-4cx^4e^{-4cx^4}=-4c$$

Then apply the Lambert W-function and you get:

$$-4cx^4=W(-4c)$$

Or $$x=\left(\frac{W(-4c)}{-4c}\right)^{1/4}$$

There are actually two real values of $W(-4c)$ when $0<c<\frac{1}{4e}$. See the linked article for details about this.

Answer 2

Another way to get an equivalent answer in terms of Lambert W function: \begin{align} \log(x) &= c x^4 \\ \log(x) x^{-4} &= c \\ \log(x^{-4}) x^{-4} &= -4c \\ \log(x^{-4}) \mathrm{e}^{\log(x^{-4})} &= -4c \\ \log(x^{-4}) &= W(-4c) \\ \log(x) &= -\frac{1}{4}W(-4c) \\ x &= \mathrm{e}^{-\frac{1}{4}W(-4c)} \end{align}

A real solution for $c<\frac{1}{4\mathrm{e}}$ is $x = \mathrm{e}^{-\frac{1}{4}W_{0}(-4c)}$, a second real solution for $0<c<\frac{1}{4\mathrm{e}}$ is $x = \mathrm{e}^{-\frac{1}{4}W_{-1}(-4c)}$. Some software systems that support $W_0$ and $W_{-1}$ are mentioned in Lambert W function:Software, for example gsl_sf_lambert_W0 and gsl_sf_lambert_Wm1 functions from the GNU Scientific Library.