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Thread: Logs Problem

  1. #1
    Newbie richard_c's Avatar
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    Logs 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
    Attached Files Attached Files
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  2. #2
    Super Member wingless's Avatar
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    Quote Originally Posted by richard_c View Post
    $\displaystyle q = K \sqrt[n]{d^m}$
    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.
    $\displaystyle q = K \sqrt[n]{d^m}$
    $\displaystyle \frac{q}{K}=d^{\frac{m}{n}}$
    $\displaystyle \left(\frac{q}{K}\right)^{\frac{n}{m}}= d$
    $\displaystyle \frac{n}{m}= \text{Log}_{ \frac{q}{K}} d$

    You can find $\displaystyle \frac{n}{m}$ for your values now.
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  3. #3
    Newbie richard_c's Avatar
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    in response

    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
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  4. #4
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    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)} $

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  5. #5
    Newbie richard_c's Avatar
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    response

    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
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by richard_c View Post
    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
    $\displaystyle \frac{m}{n} = \frac{ln \left ( \frac{q}{K} \right ) }{ln(d)}$

    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
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