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Math Help - Evaluating Logarithmics

  1. #1
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    Evaluating Logarithmics

    log_26 + log_{28} - log_23

    How do I put them into common bases? I was thinking of using power law, but there are no exponents

    Answer: 4
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Macleef View Post
    log_26 + log_{28} - log_23

    How do I put them into common bases? I was thinking of using power law, but there are no exponents

    Answer: 4
    i assume, the question should be \log_2 6 + \log_2 8 - \log_2 3

    if so, note that \log_a b^x = x \log_ab and \log a + \log b = \log ab, \log a - \log b = \log \frac{a}{b}

    thus we have \log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3
     = 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4..
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  3. #3
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    Quote Originally Posted by kalagota View Post
    i assume, the question should be \log_2 6 + \log_2 8 - \log_2 3

    if so, note that \log_a b^x = x \log_ab and \log a + \log b = \log ab, \log a - \log b = \log \frac{a}{b}

    thus we have \log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3
     = 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4..
    I think you're right, there must be a typo within the question in the textbook
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