# help to rearrang equation

• October 14th 2012, 07:03 AM
killick
help to rearrang equation
Hello

Iv been told to rearrang the following equation:

nm+1 = (Cm+1 x (V - U)m+1) / (Q x D)m+1

so that:

nm+1 x Q x D = ??????

every time iv done it iv been told i got it wrong.
• October 14th 2012, 08:06 AM
skeeter
Re: help to rearrang equation
$n^{m+1} = \frac{C^{m+1}(V-U)^{m+1}}{(QD)^{m+1}}$

$n^{m+1} = \frac{C^{m+1}(V-U)^{m+1}}{(QD)(QD)^m}$

$n^{m+1}QD = \frac{C^{m+1}(V-U)^{m+1}}{(QD)^m}$
• October 14th 2012, 09:22 AM
killick
Re: help to rearrang equation
Thanks for the help skeeter.
This there a way to get a simalar equation but without the (QD)^m factor?
• October 14th 2012, 11:21 AM
skeeter
Re: help to rearrang equation
I get the feeling there is more to this problem than what you've posted. Please state the original problem and the given directions ...
• October 14th 2012, 12:48 PM
HallsofIvy
Re: help to rearrang equation
Quote:

Originally Posted by killick
Hello

Iv been told to rearrang the following equation:

nm+1 = (Cm+1 x (V - U)m+1) / (Q x D)m+1

so that:

nm+1 x Q x D = ??????

every time iv done it iv been told i got it wrong.

I suspect you really want n x Q x D= rather than the n+1 power of r.

From $n^{m+1}= \frac{C^{m+1}(V- U)^{m+1}}{(QD)^{m+1}}$, you can write it as $n^{m+1}= \left(\frac{C(V- U)}{QD}\right)^{m+1}$
and take the n+1 root to get $n=\frac{C(V- U)}{QD}$ and then multiply both sides by QD:
$nQD= C(V- U)$
• October 17th 2012, 04:47 AM
killick
Re: help to rearrang equation
Iv got the equation:

I = n^(m+1) QD K W X

But need to substitute the n^(m+1) QD part using this equation:

n^(m+1) = (C^(m+1) x (V - U)^(m+1)) / (Q x D)^(m+1)

But with out any QD factors left on the right hand side