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    GCSE student here... just wondering, how would you solve this equation (I presume you use logarithms): 8^m = 27 or 8 to the power of m = 27
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    take logs of both sides
    so then you can take the m to the front where it is m log 8 = log 27 then you divide both sides by log 8
    so the answer is log 27/log 8 (put it into the calculator )
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    (Original post by Xez)
    GCSE student here... just wondering, how would you solve this equation (I presume you use logarithms): 8^m = 27 or 8 to the power of m = 27
    I'm going to give a full solution as you are presumably completely new to this topic.

    8^m=27

    \log 8^m = \log 27

    A law of logarithms is \log a^b = b \log a

    Consequently m \log 8 = \log 27

    Finally divide each side by \log 8
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    (Original post by Mr M)
    I'm going to give a full solution as you are presumably completely new to this topic.

    8^m=27

    \log 8^m = \log 27

    A law of logarithms is \log a^b = b \log a

    Consequently m \log 8 = \log 27

    Finally divide each side by \log 8
    Thank you! Also, presuming m is log 27 / log 8, then what is 4^m? I know it is 9, but can you explain why? So basically what is 4^(log 27/log 8)
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    (Original post by Xez)
    Thank you! Also, presuming m is log 27 / log 8, then what is 4^m? I know it is 9, but can you explain why? So basically what is 4^(log 27/log 8)
    Just put 4 to the power of the answer you just got into the calculator.

    However this is not the method you were supposed to use. You just needed to apply the laws of indices.

    8^m = 27

    (2^3)^m = 3^3

    (2^m)^3 = 3^3

    (2^m) = 3

    (2^m)^2 = 3^2

    (2^2)^m = 9

    4^m = 9
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    (Original post by Mr M)
    Just put 4 to the power of the answer you just got into the calculator.

    However this is not the method you were supposed to use.

    8^m = 27

    (2^3)^m = 3^3

    (2^m)^3 = 3^3

    (2^m) = 3

    (2^m)^2 = 3^2

    (2^2)^m = 9

    4^m = 9
    Yes, that's the question from the IMC, but I wanted to know if it was possible to do by logs without a calculator. I suppose not then? You gave the exact correct "proper" solution, so props to you for that aha.
 
 
 
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