Making T the Subject: V*T to the Power n = C

[jamesd2008]

Hi, could anyone show me how to make T the subject?

V*T to the power n = C

Thanks

 

[CompuChip]

If
T^n = c
then
(T^n)^{1/n} = T = c^{1/n}

 

[jamesd2008]

Thanks compu, could you just explain the rule as to why it is C^1/n?

Thanks James

 

[jamesd2008]

If n=0.3 Then T^3/10 so C would be the 10 root of C to the power of 3. Is that right? or am i getting my indices and power rules wrong?
Thanks for any help offerd

James

 

[timmay]

Thanks compu, could you just explain the rule as to why it is C^1/n?

Thanks James

Knowing that the logarithm of x^n is n times the logarithm of x:

VT^{n}=C

T^{n}=\frac{C}{V}

log \left[ T^{n} \right] = log \left[ \frac{C}{V} \right]

n log \left[ T \right] = log \left[ \frac{C}{V} \right]

log \left[ T \right] = \frac{1}{n} log \left[ \frac{C}{V} \right]

log \left[ T \right] = log \left[ \left( \frac{C}{V} \right)^{\frac{1}{n}} \right]

T = \left( \frac{C}{V} \right)^{\frac{1}{n}}

Many thanks to all for bracket help.

 

[jamesd2008]

Thanks Timmay. Great explanation. Just a couple of things i don't get. Why is nLog(T)=Log(c/v) then become log (T)=1/nlog(c/v). And why is not just T^n=c becomes T=nroot of c?

Thanks for all your reply's a great help James

 

[CompuChip]

Thanks Timmay. Great explanation. Just a couple of things i don't get. Why is nLog(T)=Log(c/v) then become log (T)=1/nlog(c/v).

Because if AB=C then B=C/A?
I.e. primary school algebra

And why is not just T^n=c becomes T=nroot of c?

It is. Raising something to the power 1/n is the same as taking the n-th root (cf. a^(1/2) = sqrt(a) = 2root(a)).

[Just realized what the confusion might be: note that my c is not your C... you first have to rewrite V T^n = C to T^n = c to apply what I said. ]

 

[minger]

Can't quite get the hang of brackets still, but you should be able to follow.

Try

\left( ... \right)

Rather than simply (...). \left[ \right] also works :)

 

[CompuChip]

Like this:
\left\{ \log\left[ \left(\frac{\left( C \right)}{V} \right)^{\frac{1}{n}} \right] \right\}

(click to see the source)