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

V=7c^2l / .88

Thanks I'm useless a formula

Is that what you mean?

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

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

Kriz

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

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

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