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krizzyb · May 25, 2009, 10:36 AM

how do I re arrange his formula to make c the subject

V=7c^2l / .88

Thanks I'm useless a formula

Unknown008 · May 25, 2009, 11:21 AM

Is that what you mean?

$V=\frac{7c^{2l}}{0.88}$

krizzyb · May 26, 2009, 03:32 AM

Yes that's the problem I'm sorry I couldn't type it in like that I'm useless at computing

Kriz

Perito · May 26, 2009, 05:10 AM

$V=\frac{7c^{2l}}{0.88}$

$\frac {0.88V}{7}=c^{2l}$

$log(\frac {0.88V}{7})=log(c^{2l})=2l\,log(c)$

$\frac {log(\frac {0.88V}{7})}{2l}=\frac {log(0.88)+log(V)-log(7)}{2l}=log(c)$

take the antilog

$10^{\frac {(log(0.88)+log(V)-log(7))}{2l}}=10^{log(c)}=c$

Unknown008 · May 26, 2009, 07:44 AM

Ok, I would have done it if Perito hadn't done it yet. Nice work Perito! :)