What is the solution for a general case of $ax^m = e^{b/x^n}$? What is the solution for a general case of $ax^m = e^{b/x^n}$?
I am a bit new to non elementary function, but it seems Lambert W function is a probable solution. Upon checking, it seems, it requires some form of symmetry, what if there is no known symmetry? How do we solve this analytically?
 A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align} 
a\,x^m &= \exp\Big(\frac b{x^n}\Big)
\end{align} 
Let $y=\ln x$. Then we have
\begin{align}
\ln(a)+m\,y&=b\,\exp(-n\,y)
,\\
\frac nm\,\ln(a)+n\,y &= \frac nm\,b\exp(-n\,y)
,\\
\Big(\frac nm\,\ln(a)+n\,y \Big)\,\exp(n\,y) &= \frac nm\,b
,\\
\Big(\frac nm\,\ln(a)+n\,y \Big)\,\exp(n\,y)\,\exp\Big(\frac nm\,\ln(a)\Big) 
&= \frac nm\,b\,\exp\Big(\frac nm\,\ln(a)\Big)
,\\
,\\
\Big(\frac nm\,\ln(a)+n\,y \Big)\,\exp(\frac nm\,\ln(a)+n\,y)
&=\frac{n}m\,b\,a^{n/m}
,\\
\end{align} 
Applying the Lambert $\W$ function,
\begin{align}
\W\left(\Big(\frac nm\,\ln(a)+n\,y \Big)\,\exp(\frac nm\,\ln(a)+n\,y)\right)
&=\W\left(\frac{n}m\,b\,a^{n/m}\right)
,
\end{align}
\begin{align}
\frac nm\,\ln(a)+n\,y
&=\W\left(\frac{n}m\,b\,a^{n/m}\right)
,\\
n\,y
&=\W\left(\frac{n}m\,b\,a^{n/m}\right)-\frac nm\,\ln(a)
,\\
y
&=\frac1n\,\W\left(\frac{n}m\,b\,a^{n/m}\right) +\ln(a^{-1/m})
,\\
x&=
a^{-1/m}\,\exp\left(\frac1n\,\W\left(\frac{n}m\,b\,a^{n/m}\right) \right)
.
\end{align} 
The analysis of the argument of $\W$ gives the number of real solutions:
\begin{align}
t=\frac{n}m\,b\,a^{n/m}
:
\begin{cases}
t<-\frac1{\e}\Longrightarrow \text{no real solutions}
,\\
t=-\frac1{\e} \text{ or }
t\ge0 \Longrightarrow \text{one real solution, use }\Wp(t)
,\\
-\frac1{\e} <t<0 \Longrightarrow \text{two real solutions, use } \Wp(t) \text{ and }\Wm(t)
\end{cases}
.
\end{align}
Note that this result is exactly the same 
as in the other answer:
\begin{align}
&\phantom{=}a^{-1/m}\,\exp\left(\frac1n\,\W\left(\frac{n}m\,b\,a^{n/m}\right) \right)
\\
&=\sqrt[n]{a^{-n/m}\,\exp\left(\W\left(\frac{n}m\,b\,a^{n/m}\right) \right)}
\\
&=\sqrt[n]{
\frac{a^{-n/m}\,\frac{n}m\,b\,a^{n/m}}
{\W\left(\frac{n}m\,b\,a^{n/m}\right)}
}
\\
&=\sqrt[n]{
\frac{\frac{n}m\,b}
{\W\left(\frac{n}m\,b\,a^{n/m}\right)}
}
.
\end{align}
$\endgroup$
A: Let $t:=\dfrac b{x^n}$ and the equation is
$$a\left(\frac bt\right)^{m/n}=e^t$$ or
$$a^{n/m}\frac bt=e^{nt/m}.$$
Now with $s:=\dfrac{nt}m$,
$$a^{n/m}b\frac nm=se^{s}.$$
Finally,
$$x=\sqrt[n]{\frac{nb}{mW\left(a^{n/m}b\dfrac nm\right)}}$$
