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Math Help - logarithims question, any help appreaciated

  1. #1
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    logarithims question, any help appreaciated

    I am stuck with a logarithims problem that i just can't seem to solve.
    The problem is,

    Determine x in

    10 to power 2x = 5e to power 2x + 2

    The two different bases are confusing me, can any one offer any suggestions on how to go about solving this? am i missing something obvious?

    Any help would be appreaciated
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  2. #2
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    Quote Originally Posted by paul90900 View Post
    I am stuck with a logarithims problem that i just can't seem to solve.
    The problem is,

    Determine x in

    10 to power 2x = 5e to power 2x + 2

    The two different bases are confusing me, can any one offer any suggestions on how to go about solving this? am i missing something obvious?

    Any help would be appreaciated
    10^{2x} = 5e^{2x+2}

    natural log of both sides ...

    \ln(10^{2x}) = \ln(5e^{2x+2})

    2x \cdot \ln(10) = \ln(5) + 2x + 2

    2x \cdot \ln(10) - 2x = \ln(5) + 2

    2x[\ln(10) - 1] = \ln(5) + 2

    x = \frac{\ln(5)+2}{2[\ln(10)-1]}
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  3. #3
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    It works but i am still a bit confused

    Many thanks for the reply but i must admit i'm a bit confused with the line



    why is the 2x + 2 added to ln(5)

    Sorry if this is a stupid question,
    i have only just started up a maths course after 10 years away from education.
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  4. #4
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    \ln(10^{2x}) = \ln(5e^{2x+2})

    2x \cdot \ln(10) = \ln(5) + \ln(e^{2x + 2})


    note that \ln(e^a) = a
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  5. #5
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    many thanks, i see it now
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