[FONT='Calibri','sans-serif']can anyone help me with the following:[/FONT]

[FONT='Calibri','sans-serif']i have an equation - as attachment[/FONT]

i need to find the ratio n/m

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- Dec 10th 2007, 12:07 PMrichard_cLogs Problem
[FONT='Calibri','sans-serif']can anyone help me with the following:[/FONT]

[FONT='Calibri','sans-serif']i have an equation - as attachment[/FONT]

i need to find the ratio n/m - Dec 10th 2007, 12:28 PMwingless
I see

*litre*and*mm*there.I'm not sure whether it needs, but first you have to equalize the units. Then you can solve the equation for n/m.

$\displaystyle q = K \sqrt[n]{d^m}$

$\displaystyle \frac{q}{K}=d^{\frac{m}{n}}$

$\displaystyle \left(\frac{q}{K}\right)^{\frac{n}{m}}= d$

$\displaystyle \frac{n}{m}= \text{Log}_{ \frac{q}{K}} d$

You can find $\displaystyle \frac{n}{m}$ for your values now. - Dec 10th 2007, 02:40 PMrichard_cin response
Hi WIngless,

Thanks very much for your help.

May I ask a question, as I dont understand maths that easy.

http://www.mathhelpforum.com/math-he...54e37260-1.gif

http://www.mathhelpforum.com/math-he...74d6b77d-1.gif

in the transition between the two top equations, why is it that the power can just be moved across? Im not doubting you, I just want to learn why.

THanks - Dec 10th 2007, 06:59 PMSoroban
Hello, Richard!

Quote:

$\displaystyle q \;=\;K\sqrt[n]{d^m}$

Find the ratio $\displaystyle \frac{m}{n}$

We have: .$\displaystyle K\sqrt[n]{d^m}\quad\Rightarrow\quad K\!\cdot\!d^{\frac{m}{n}}\;=\;q\quad\Rightarrow\qu ad d^{\frac{m}{n}} \:=\:\frac{q}{K}$

Take logs: .$\displaystyle \ln\left(d^{\frac{m}{n}}\right) \;=\;\ln\left(\frac{q}{K}\right)\quad\Rightarrow\q uad \frac{m}{n}\!\cdot\!\ln(d) \;=\;\ln\left(\frac{q}{K}\right) $

Therefore: .$\displaystyle \frac{m}{n} \;=\;\frac{\ln\left(\frac{q}{K}\right)}{\ln(d)} $

- Dec 11th 2007, 04:17 AMrichard_cresponse
Hi,

When i punch the figures into the top equation, the figure comes out as -550!??!

I cant get the bottom equation to work, at all! Can someone help me out?

Thanks - Dec 11th 2007, 06:41 AMtopsquark
What top and what bottom equation? The only one I can think to help you with is

$\displaystyle \frac{m}{n} = \frac{ln \left ( \frac{q}{K} \right ) }{ln(d)}$

Using the numbers in the units you gave:

$\displaystyle \frac{m}{n} = \frac{ln \left ( \frac{5.51}{32 \times 10^{-6}} \right ) }{ln(91.9)}$

$\displaystyle \frac{m}{n} = \frac{ln(172188 ) }{ln(91.9)} = \frac{12.0563}{4.5207} = 2.66692$

-Dan