Solve for $x$ in $v×b=1-(\sqrt[r]{b-(v×b)}×\dfrac{2^{a/2}}{\sqrt[r]{(1-x)×v}})$

This one is over my head (and causes an error in the best online calculator I could find and used previously).

$v×b=1-(\sqrt[r]{b-(v×b)}×\dfrac{2^{a/2}}{\sqrt[r]{(1-x)×v}})$

I need to know the formula for x. If needed, $a = 3$ and $b=0.5$.

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We do it one step at a time. It would be easy to combine steps. And we will do a truly silly thing. Note that your expression has a factor of $\sqrt[r]{v}$ in the numerator and in the denominator. Of course they cancel, and we really should cancel them immediately. But we will wait.

We start with your $v×b=1-(\sqrt[r]{v}×\dfrac{2^{a/2}}{\sqrt[r]{(1-x)×v}})$.

Edit: This was the equation at the time the answer was being typed. The equation was later changed. We do not solve the changed problem, it involves similar moves.

The first thing to do is to remove the annoying multiplication signs, and the unnecessary parenthesis: $$vb=1-\sqrt[r]{v}\frac{2^{a/2}}{\sqrt[r]{(1-x)v}}.$$ Subtract $1$ from both sides. We get $$vb-1=-\sqrt[r]{v}\frac{2^{a/2}}{\sqrt[r]{(1-x)v}}.$$ Multiply by $-1$, that is, change signs. We get $$1-vb=\sqrt[r]{v}\frac{2^{a/2}}{\sqrt[r]{(1-x)v}}.$$ Do minor cleaning: $$1-vb=\frac{2^{a/2}\sqrt[r]{v}}{\sqrt[r]{(1-x)v}}.$$ Multiply both sides by $\sqrt[r]{(1-x)v}$: $$(1-vb)\sqrt[r]{(1-x)v}=2^{a/2}\sqrt[r]{v}.$$ Divide both sides by $1-vb$: $$\sqrt[r]{(1-x)v}=\frac{2^{a/2}\sqrt[r]{v}}{1-vb}.$$ Take the $r$-th power of both sides. Note that the $r$-th power of $2^{a/2}$ can be written simply as $2^{ar/2}$. Things change in appearance quite a bit: $$(1-x)v=\frac{2^{ar/2}v}{(1-vb)^r}.$$ Now we do something that could have been done long long ago. Cancel the multiplier $v$ from both sides: $$1-x=\frac{2^{ar/2}}{(1-vb)^r}.$$ Finally, we get $$x=1-\frac{2^{ar/2}}{(1-vb)^r}.$$

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thank you very much. At the time you were answering this, I realised I left a small part out of the original equation. I will follow your steps to correct my error. Thanks again. –  IanC Jun 27 '12 at 13:57
Sorry about the annoying × and ()... just being clear ;) –  IanC Jun 27 '12 at 13:58
@IanC: Will leave the new calculation to you. The steps are similar. There will again be some cancellation, less dramatic. It is just a process of unburying $x$. –  André Nicolas Jun 27 '12 at 14:17
I worked it out fine. Thanks again for your very clear answer. –  IanC Jun 27 '12 at 14:19