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Math Help - solve the logarithm equation

  1. #1
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    Question solve the logarithm equation

    im not to sure how far i should be going to solve the equation, am i supposed to find out what x is. if so this is what i have come up with
    Question: 5^x-1=2^3x-1 give answer to 3s.f

    Workings
    log5 ^ 23x-1=x-1
    3x-1log5^2=x-1
    log5^2=x-1/3x-1
    1.4=x-1/3x-1
    1.4(3x-1)=x-1
    4.2x-1.4=x-1
    3.2x-1.4=-1
    3.2x=0.4
    x=0.4/3.2
    x=0.125
    3.s.f= 0.13
    i know somewhere i have gone really wrong, but not quite sure where. any tips?
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  2. #2
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    5^x-1=2^3x-1

    (x-1)ln 5 = (3x-1) ln 3

    = 1.60943791x - 1.60943791 = 3.29583687x- 1.09861229

    1.68639895x = - 0.510825624

    x = -0.317393806
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  3. #3
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    Exclamation

    (x-1)ln 5 = (3x-1) ln 3
    where you have done this part here
    why would it be (3x-1)ln3
    and not (3x-1)ln 2
    i may have missed something but would you be kind enough to give a short explanation
    thankyou
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  4. #4
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    Oh, I'm sorry. I misread that.

    Follow the steps I did except use ln 2 instead of ln 3
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  5. #5
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    Quote Originally Posted by satx View Post
    5^x-1=2^3x-1

    (x-1)ln 5 = (3x-1) ln 3

    = 1.60943791x - 1.60943791 = 3.29583687x- 1.09861229

    1.68639895x = - 0.510825624

    x = -0.317393806
    Unless your calculator is very old it's better to only get a decimal approximation at the end.


    (x-1) \ln (5) = (3x-1) \ln 3

    x \ln (5) - \ln(5) = 3x \ln (3) - \ln (3)

    x[\ln (5) - 3 \ln (3)] = \ln (5) - \ln (3)

    x = \frac{\ln (5) - \ln (3)}{\ln (5) - 3 \ln (3)}

    From what I can guess it's the same answer, just that one is exact
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