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Math Help - Condensing Logarithms

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    Condensing Logarithms

    Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
    3log_4a+5log_4b-log_4c

    please help
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  2. #2
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    Quote Originally Posted by jvrobi167 View Post
    Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
    3log_4a+5log_4b-log_4c

    please help
    Log questions are pre-calc not trig


    Either way use the power law and the addition/subtraction laws.

    Power Law: k\log_b(a) = \log_b(a^k)

    Addition Law: \log_b(a) + \log_b(c) = \log_b(ac)

    Subtraction Law: \log_b(a) - \log_b(c) = \log_b \left(\frac{a}{c}\right)


    For example \log_2(3) + 3\log_2(6) = \log_2(648)

    ( 648 = 6^3 \times 2 in case you were wondering)
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  3. #3
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    Quote Originally Posted by jvrobi167 View Post
    Okay so I am totally confused on how to do this prroblem it asking to condense the following problem:
    3log_4a+5log_4b-log_4c

    please help
    To "condense" a log expression like this, use the following "laws of logarithms"...

    log_xy^k=klog_xy

    log_xa+log_xb=log_x(ab)

    log_xa-log_xb=log_x\left(\frac{a}{b}\right)

    Hence,

    3log_4a+5log_4b-log_4c=log_4a^3+log_4b^5-log_4c

    =log_4\left(a^3b^5\right)-log_4c

    Try to finish using the law of subtraction.
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