Solving for $x$ in the equation $xa^x = y$

Question

I am trying to solve the equation

$$xa^x = y,$$

for $x$ where $x$ should be positive. The only thing known is $a < 1$. I tried taking logarithm of both side but it doesn't really lead anywhere.

Answer 1

Let $a=e^{ \alpha} $ and multiply by $\alpha$ \begin{eqnarray*} \alpha x e^{ \alpha x} =\alpha y .\\ \end{eqnarray*} Now recall the Lambert $W$ function is defined by $we^w=z$ gives $w=W(z)$. So we have \begin{eqnarray*} x&=& \frac{1}{\alpha} (W(\alpha y)) \\ x&=& \frac{1}{\ln(a)} (W(y \ln(a))). \\ \end{eqnarray*}

Answer 2

Multiply by $\log a$ to get

$$x\log a\,e^{x\log a}=y\log a$$

and invert by

$$x\log a=W(y\log a).$$