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    Write these as a single logarithm:-

    c) 5\log_aa + \left(\frac{1}{3}\right)\log_a27  + \log_a2

    The book has provided the solution of \log_a6a^5

    But I'm uncertain as to whether my solution is also valid:-

    5\log_aa + \left(\frac{1}{3}\right)\log_a27  + \log_a2 = \log_aa^5 + log_a3 + log_a2 =  \log_a6 + 5

    Many thanks.
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    (Original post by supreme)
    Write these as a single logarithm:-

    c) 5\log_aa + \left(\frac{1}{3}\right)\log_a27  + \log_a2

    The book has provided the solution of \log_a6a^5

    But I'm uncertain as to whether my solution is also valid:-

    5\log_aa + \left(\frac{1}{3}\right)\log_a27  + \log_a2 = \log_aa^5 + log_a3 + log_a2 =  \log_a6 + 5

    Many thanks.
    It's correct; but it's not a single logarithm. So false.
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    A single logarithm means just one term, with Log in it.

    [Let Log = Log to the base a... Because I'm on mobile, it makes it easier]

    First, use the power rule.
    xLogy = Logy^x

    So:
    5Loga = Loga^5
    (1/3)Log27 = Log27^(1/3) = Log3

    You now have:
    Loga^5 + Log3 + Log2

    Now, use the product rule
    Logx + Logy = Logxy

    So:
    Log(a^5x3x2)

    You now have:
    Log(6a^5)
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    (Original post by Jamdroid)
    A single logarithm means just one term, with Log in it.

    [Let Log = Log to the base a... Because I'm on mobile, it makes it easier]

    First, use the power rule.
    xLogy = Logy^x

    So:
    5Loga = Loga^5
    (1/3)Log27 = Log27^(1/3) = Log3

    You now have:
    Loga^5 + Log3 + Log2

    Now, use the product rule
    Logx + Logy = Logxy

    So:
    Log(a^5x3x2)

    You now have:
    Log(6a^5)
    Thanks
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    (Original post by multiratiunculae)
    It's correct; but it's not a single logarithm. So false.
    Yes, just shows I didn't read the question properly.
 
 
 
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