Logs Problem

**richard_c** · December 10th 2007, 12:07 PM

[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

**wingless** · December 10th 2007, 12:28 PM

Originally Posted by richard_c

$q = K \sqrt[n]{d^m}$
The equation above an equation for the discharge flowrate

K=32x 10-6
q= 5.5l/s
d=91.9mm

I am trying to find the ratio n/m

Do i take my n over d to the m and change it to d ?
If so what is the next step?

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.
$q = K \sqrt[n]{d^m}$
$\frac{q}{K}=d^{\frac{m}{n}}$
$\left(\frac{q}{K}\right)^{\frac{n}{m}}= d$
$\frac{n}{m}= \text{Log}_{ \frac{q}{K}} d$

You can find $\frac{n}{m}$ for your values now.

**richard_c** · December 10th 2007, 02:40 PM

Hi WIngless,
Thanks very much for your help.

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

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

**Soroban** · December 10th 2007, 06:59 PM

Hello, Richard!

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

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

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

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

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

**richard_c** · December 11th 2007, 04:17 AM

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

**topsquark** · December 11th 2007, 06:41 AM

Originally Posted by richard_c

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

What top and what bottom equation? The only one I can think to help you with is
$\frac{m}{n} = \frac{ln \left ( \frac{q}{K} \right ) }{ln(d)}$

Using the numbers in the units you gave:
$\frac{m}{n} = \frac{ln \left ( \frac{5.51}{32 \times 10^{-6}} \right ) }{ln(91.9)}$

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

-Dan

Thread: Logs Problem

LinkBack

Thread Tools

Display

Logs Problem

in response

response

Similar Math Help Forum Discussions

[SOLVED] Logs problem

Few HW problems involving Logs, Natural logs, and Continuity.

logs problem

logs problem

Problem with logs

Search Tags