how to solve $\log{x}=cx^4$ for $x$ I was wondering if there is a general solution for this form of equations:
$$\log{x}=cx^4$$
Tried: $$ x = e^{cx^4}\\
xe^{-cx^4}=1$$
 A: 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.
A: 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.
