[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
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.
Hello, Richard!
$\displaystyle q \;=\;K\sqrt[n]{d^m}$
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)} $
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