I've been looking at this equation for some time now:

$$e^x = yx^n$$

where $n \in \mathbb{N}_+$.

I found I can transform it to:

$$e^x = yx^n$$ $$xe^x = yx^{n+1}$$ $$x=W(yx^{n+1})$$

Or to this:

$$e^x = yx^n$$ $$x = \ln(yx^n)$$

If $yx^n$ is positive:

$$x = \ln(x^n)+\ln(y)$$

And if both $x^n$ and $y$ are positive:

$$x = n\ln(x)+\ln(y)$$

But I didn't find anyway to solve these for $x$ from there.

  • 2
    $\begingroup$ Hint: take $n$th root of both sides. $\endgroup$ – Wojowu Nov 4 '16 at 14:53

$$ yx^n=e^x $$ First, divide by $ye^x$ $$ x^n\frac{1}{e^x}=\frac 1y $$ Then, take the $n$-th root, as suggested by Wojowu in comments: $$xe^{-\frac xn}=y^{-\frac 1n}$$

Now, multiply by $-\dfrac 1n$ in order to use $W(x)e^{W(x)}=x$: \begin{align*}-\frac xne^{-\frac xn}&=-\frac{y^{-\frac 1n}}{n} \\ -\frac{x}{n}&=\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right) \\ x&=-n\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right) \end{align*} and the equation is solved.


You can not solve the above equation explicit in the form $x=f(y)$


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