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Math Help - Simple Log problem

  1. #1
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    Simple Log problem

    Solve algebraically using logs.

    2^x = 5^x+1

    So far I've got x log 2 = (x+1) log 5

    I think I'm going wrong on the next step. Am I to start isolating X by dividing log 2 by log 5 or is there something I'm missing?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by cantstop01 View Post
    Solve algebraically using logs.

    2^x = 5^x+1

    So far I've got x log 2 = (x+1) log 5

    you only do this when the question is 2^x=5^{x+1}

    I think I'm going wrong on the next step. Am I to start isolating X by dividing log 2 by log 5 or is there something I'm missing?
    .
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  3. #3
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    i am don't think it is right, but in this specific problem its true...
    2^x=5^x+1
    2^x-5^x=1  /\times\log
    \log(2^x)-\log(5^x)=\log(1)
    x\times\log(2)-x\times\log(5)=\log(1)
    x=\frac{0}{(\log(2)-\log(5))}
    x=0
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