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Math Help - Re-arranging formula

  1. #1
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    Re-arranging formula

    Hi, im new here. Just wondering if someone could walk me through solving this for equation for V2. I have a solution but some parts seemed to have been skipped and i dont quite understand why:

    150*V2^1.3 = 100*((5*10^-3)+V2)^1.3

    Thanks
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  2. #2
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    150V_2^{1.3} = 100(V_2 + 5 \cdot 10^{-3})^{1.3}

    Take the log of both sides (I shall use natural log)

    \ln(150V_2^{1.3}) = \ln[100(V_2 + 5 \cdot 10^{-3})^{1.3}]

    You can then use the laws of logs to separate what's inside

    \ln(150) + 1.3\ln(V_2) = \ln(100) + 1.3\ln(V_2+5 \cdot 10^{-3})


    1.3[\ln(V_2+5 \cdot 10^{-3}) - \ln(V_2)] = \ln(150) - \ln(100) = \ln(1.5)

    \ln \left(\dfrac{V_2+5 \cdot 10^{-3}}{V_2}\right) = \dfrac{\ln(1.5)}{1.3}


    Chances are there is an easier way than using logs but there is more than one way to skin a mathematical cat
    Last edited by e^(i*pi); January 13th 2011 at 03:42 AM. Reason: forgot to divide by 1.3 on the RHS
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  3. #3
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    \displaystyle 150V_2^{\frac{13}{10}} = 100(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}

    \displaystyle \frac{150}{100} = \frac{(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}}{V_2^{\frac{13}{10}}}

    \displaystyle \frac{3}{2} = \left(\frac{5\cdot 10^{-3} + V_2}{V_2}\right)^{\frac{13}{10}}

    \displaystyle \left(\frac{3}{2}\right)^{\frac{10}{13}} = \frac{5\cdot 10^{-3} + V_2}{V_2}

    \displaystyle \frac{3^{\frac{10}{13}}}{2^{\frac{10}{13}}} = \frac{5\cdot 10^{-3}}{V_2} + 1


    Can you go from here?
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  4. #4
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    I thought i could but im not getting the same answer as my solution.

    Logs were never my strong point...
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  5. #5
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    You don't need logarithms, see my post above...
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  6. #6
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    Yeah, i just worked through it compared to the given solution. It has been done the same way as you "pove it" only difference is your (3^10/13)/(2^10/13) is written as (150/100)^1/1.3. Thanks for the help both of you, i can make sense of it now
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  7. #7
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    I tell a lie... I'm not 100% sure why the +1 gets rid of the V2?!?
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  8. #8
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    Prove it split the fraction into two pieces. More precisely he used the following:

    \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}.

    Do you see it now?
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  9. #9
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    I do... spot on that, cheers
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