What is the solution for a general case of $ax^m = e^{b/x^n}$?

Question

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?

Answer 1

$\require{begingroup} \begingroup$

$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\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}

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Answer 2

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,

$$x=\sqrt[n]{\frac{nb}{mW\left(a^{n/m}b\dfrac nm\right)}}$$