re-arrange equation $L=2^{10(v-1)} v^2$

Question

Is it possible to re-arrange this equation to make v the subject?

$$L=v^2 . 2^{10(v-1)}$$

If so, what is the answer?

If it helps (which by excluding zero it should)... $$0<v<1$$

I have tried pages of scribbling and got nowhere. In desperation I have tried solving each product separately (easy enough), and then tried to get the right overall result from some combination of $sqrt(L)$ and $(log2(L)+10)/10$ but that hasn't got me anywhere.

I am out of my depth. Please help.

Answer 1

If make v the subject means solve for $v$, there are solutions in terms of the LambertW function. With Maple I get

$$v = \frac{1}{5}\frac{W\big(\pm160\ln 2 \sqrt{L}\big)}{\ln 2}$$

For the ranges $0 < v,L < 1$ you have to use the $+$ sign, the $-$ will give complex $v$.

You can also use Wolfram Alpha with the input solve 2^(10*(v-1))*v^2 = L (they call $W$ the product log function).