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Math Help - Logarithm Problems

  1. #1
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    Logarithm Problems

    How would one solve for x in the following, using the laws of logarithms?

    1) 5^x+3^2x=92

    2) 4*5^x-3*0.4^2x=11

    In 1), this is what I've done so far:

    5^x+3^2x=92
    log(5^x+3^2x)=log 92
    log(5+3^2)^x=log 92
    x log 14=log 92

    but I don't think that it's correct, and things aren't any clearer for question 2).
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  2. #2
    MHF Contributor
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    Hello DemonX01
    Quote Originally Posted by DemonX01 View Post
    How would one solve for x in the following, using the laws of logarithms?

    1) 5^x+3^2x=92

    2) 4*5^x-3*0.4^2x=11

    In 1), this is what I've done so far:

    5^x+3^2x=92
    log(5^x+3^2x)=log 92
    log(5+3^2)^x=log 92
    x log 14=log 92

    but I don't think that it's correct, and things aren't any clearer for question 2).
    For number (1), you're right: your working is not correct.

    However, I don't know what method you are supposed to use to solve these equations. The only way I can see is to use some sort of approximate numerical method. Doing this, I can tell you that the answers are: (1) 1.9311, and (2) 0.6758, correct to 4 d.p.

    But I cheated a bit and used a spreadsheet.

    Grandad
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  3. #3
    Super Member Quacky's Avatar
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    Nov 2009
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    Windsor, South-East England
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    Remember that 3^{2x}=(3^2)^x
    So you actually have:
    5^x+9^x=92

    I'll edit this post with an answer if I find it, but as Grandad couldn't, I doubt I'll be able to take it any further without using a spreadsheet, as I've next to no maths knowledge whatsoever by comparison.
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