Am I going to use the Lambert W function for this one?

Question

The question is simple. Now I have:

$a \cdot \ln(b)=p$

and

$a\cdot b^3=q$

Can I make $a$ and $b$ the subjects and express them in terms of $p$ and $q$?

I looked up a bit and seems that the Lambert W function, $z=W(ze^z)$ is what is relevant, but I still cannot manage to make $a$ and $b$ the subjects.

Thanks in advance!

Answer 1

I'd let $b=e^c$ and then divide the equations:

$$\frac{\ln e^c}{e^{3c}} = \frac{p}{q}.$$

So that

$$ce^{-3c} = \frac{p}{q}$$

or

$$-3ce^{-3c} = -3\frac{p}{q}.$$

Then hit both sides with $W$

$$-3c = W(-3p/q)$$

$$\ln b = -\frac{1}{3} W(-3p/q)$$

So

$$b = \exp(-\frac{1}{3} W(-3p/q)).$$

Then $a$ follows.

Answer 2

For the equations \begin{align} a \, \ln(b) &= p \\ a \, b^{n} &= q \end{align} consider the following. From the first equation $b = e^{p/a}$ and then, with use of the Lambert W-function, \begin{align} a \, e^{n p/a} &= q \\ \frac{1}{a} \, e^{- n p/a} &= \frac{1}{q} \\ - \frac{n p}{a} \, e^{- n p/a} &= - \frac{n p}{q} \\ - \frac{n p}{a} &= W\left(- \frac{n p}{q} \right) \\ \frac{p}{a} &= - \frac{1}{n} \, W\left(- \frac{n p}{q} \right). \end{align} Now, $$b = Exp\left[- \frac{1}{n} \, W\left(- \frac{n p}{q} \right) \right].$$