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Math Help - rearranging power and log equations

  1. #1
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    rearranging power and log equations

    Hello,

    I have two equations that I want to reaarange, one a power, the other a log. I can solve the orignal equations but am getting stuck trying to make x the subject. Any help would be appreciated.

    eq. 1: y= 4.3316x0.4671

    To solve this (for x=8) I did the following: 4.3316 * 80.4671
    = 8^(1/10 000) PWR 4671
    = 4.3316 *2.64
    = 11.44
    I think this is right but don't know how to rearrange to get x= ?

    Eq 2. y=9.5818ln(x) - 9.5878
    Can solve this ok but stuck on same point - getting x=

    Thanks
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: rearranging power and log equations

    Eq 2:
    y=9,5818\ln(x)-9,5878 \Leftrightarrow y+9,5878=9,5818\ln(x) \Leftrightarrow \frac{y+9,5878}{9,5818}=\ln(x) \Leftrightarrow x=e^{\frac{y+9,5878}{9,5818}}

    Eq 1:
    This one is not clear in my opinion.
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  3. #3
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    Re: rearranging power and log equations

    I assume the problem (1) is to solve y= 4.3316x^{0.4671} for x. (If you don't use LaTeX, at least use "^" to indicate powers.)

    Start with the obvious: divide both sides by 4.3316: \frac{y}{4.3316}= x^{0.4671}
    and then take the .4671 root of both sides: x= \left(\frac{y}{4.3316}\right)^{1/0.4671}.

    No logarithm need.

    If the x had been in the exponent, then you would need the logarithm:
    y= 4.3316(0.4671)^x
    \frac{y}{4.3316}= (0.4671)^x
    log\left(\frac{y}{4.3316}\right)= log(0.4671^x)= xlog(0.4671)
    x= \frac{log\left(\frac{y}{4.43316}\right)}{log(0.467  1)}
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  4. #4
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    Re: rearranging power and log equations

    Thanks for both your answers - it's so obvious when someone shows you how!

    I also apologise for the rather sloppy nature of the post re. layout etc. Will make sure I'm clear if I need to post again.

    Thanks again.
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  5. #5
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    Re: rearranging power and log equations

    [QUOTE=Siron;675593]Eq 2:
    y=9,5818\ln(x)-9,5878 \Leftrightarrow y+9,5878=9,5818\ln(x) \Leftrightarrow \frac{y+9,5878}{9,5818}=\ln(x) \Leftrightarrow x=e^{\frac{y+9,5878}{9,5818}}

    Sorry to be stupid but I can't seem to get a sensible answer drom the above equation that you kindly showed me how to rearrange. When I solve the original equation for y using a series of values for x, I can't then use those same values of y to get the corresponding values of x that I used - I end up with insanely large values for x.

    What I would give to be good at maths
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: rearranging power and log equations

    I don't get your point, do you understand the steps I did or not? Can you give an example where you're stuck?

    And it's not stupid, it's important to ask things what you don't understand.
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  7. #7
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    Re: rearranging power and log equations

    Ok. Here is my problem. Let's say in the original equation x =49. If I am correct this would give a value for y of 27.7. So now I want to use the rearranged equation, using this value of y (27.7), to get back to the original value of x I used (49) - to check I understand how to do it. This is where I am stuck. Looking at the equation, e is the base, which is ln or 2.718281828. Am I right here? y+9.5878/9.5818 is the power? This produces absurdly large anwers. I'm sure there is something pretty fundamental that I am not understanding?!
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: rearranging power and log equations

    It's the same, indeed when x=49 then y\approx 27,7028. When we substitutie this value for y in the equation:
    x=e^{\frac{y+9,5878}{9,5818}} then we get:
    x=e^{\frac{27,7028+9,5878}{9,5818}}
    \Leftrightarrow x=48,998\approx 49

    So the rearranged equation is fully equivalent with the original one.
    Does this answer your question?
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  9. #9
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    Re: rearranging power and log equations

    Yes thanks. Just realised what I've been doing wrong -trying to do the sum in the wrong order, which explains why I was getting it wrong and now it is correct. Thanks again.
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: rearranging power and log equations

    You're welcome!
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