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Math Help - Solving Logarithms

  1. #1
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    Solving Logarithms

    How would I go about solving for "x" for something like this (should be fairly simple):

    log7^x=log7^9

    edit: thats supposed to be base 7. how would you enter the code for that?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cjmac87 View Post
    How would I go about solving for "x" for something like this (should be fairly simple):

    log7^x=log7^9
    if log(something) = log(something) then the somethings must be equal, the log function is one-to-one.

    so just drop the logs. you have 7^x = 7^9, now what is x?
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  3. #3
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    thanks that helps.

    ok what about something thats not one-to-one:
    log_2{2^-1}=x
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cjmac87 View Post
    thanks that helps.

    ok what about something thats not one-to-one:
    log_2{2^-1}=x
    you should know that \log_aa^n = n
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  5. #5
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    ah, oh course. very simple.

    one last thing i need cleared up which is probably easy to:

    log_2{16} can be solved how?
    how about log_{16}{1/4}?
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by cjmac87 View Post
    ah, oh course. very simple.

    one last thing i need cleared up which is probably easy to:

    log_2{16} can be solved how?
    how about log_{16}{1/4}?
    let \, \log_2 {16} = x

    then

    2^x = 16 = 2^4
    now, you can continue..

    the other one, note that

    \log_a {\frac{c}{d}} = \log_a {c} - \log_a {d}

    and \, \log_a 1 = 0 \, for \, all \, a>0

    can you do the rest?
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  7. #7
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    You seem to be missing an important concept:

    Logarithm = Exponent

    Trust me on this. What 2 roles are played by the 4, below?

    log_{2}(16)\;=\;4\;\iff\;2^{4}\;=\;16

    You do the other one.
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