Danika Mckay

2022-10-24

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

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.

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.

latatuy

Beginner2022-10-25Added 12 answers

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

$$v=\frac{1}{5}\frac{W{\textstyle (}\pm 160\mathrm{ln}2\sqrt{L}{\textstyle )}}{\mathrm{ln}2}$$

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 $${2}^{10\cdot (v-1)}\cdot {v}^{2}=L$$ (they call W the product log function).

$$v=\frac{1}{5}\frac{W{\textstyle (}\pm 160\mathrm{ln}2\sqrt{L}{\textstyle )}}{\mathrm{ln}2}$$

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 $${2}^{10\cdot (v-1)}\cdot {v}^{2}=L$$ (they call W the product log function).

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