# re-arrange equation L=2^(10(v-1)) v^2

re-arrange equation $L={2}^{10\left(v-1\right)}{v}^{2}$
Is it possible to re-arrange this equation to make v the subject?
$L={v}^{2}.{2}^{10\left(v-1\right)}$
If so, what is the answer?
If it helps (which by excluding zero it should)...
$0
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 $\left(log2\left(L\right)+10\right)/10$ but that hasn't got me anywhere.
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latatuy
If make v the subject means solve for $v$, there are solutions in terms of the LambertW function. With Maple I get
$v=\frac{1}{5}\frac{W\left(±160\mathrm{ln}2\sqrt{L}\right)}{\mathrm{ln}2}$
For the ranges $0 you have to use the $+$ sign, the $-$ will give complex $v$.
You can also use Wolfram Alpha with the input solve ${2}^{10\cdot \left(v-1\right)}\cdot {v}^{2}=L$ (they call W the product log function).