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Math Help - Logs and Exponents

  1. #1
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    Logs and Exponents

    I'm a little stumped here, as we have a constant to the power of an exponent.. How would I simplify this:

    6^{\log_{2}n}

    --------------------------------------------------------------------------

    I would also appreciate it if someone could tell me if the answer I came up with for the following is correct:
    Simplify the following:
    \log_{10}(10(\log_{10}n^3)100^{2n})

    My answer:
    \log_{10}(30\log_{10}n) + 4n

    I feel my answer isn't complete as I'm not too sure how to further simplify what's left of the addition operator.
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  2. #2
    Senior Member vincisonfire's Avatar
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    For the first one you have
     6^{log_2 n} = (2\cdot 3)^{log_2 n} = 2^{log_2 n} 3^{log_2 n} = n 3^{log_2 n}
    For the second
     \log_{10}(30\log_{10}n) + 4n = \log_{10}^2(n) + \log_{10}(3) + \log_{10}(10) + 4n = \log_{10}^2(n) + \log_{10}(3) + 1 + 4n
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  3. #3
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    Quote Originally Posted by vincisonfire View Post
    For the first one you have
     6^{log_2 n} = (2\cdot 3)^{log_2 n} = 2^{log_2 n} 3^{log_2 n} = n 3^{log_2 n}
    For the second
     \log_{10}(30\log_{10}n) + 4n = \log_{10}^2(n) + \log_{10}(3) + \log_{10}(10) + 4n = \log_{10}^2(n) + \log_{10}(3) + 1 + 4n
    Alternatively:

     \log_{2}{n} = \log_{6}n \times \log_{2}6

    Hence:

     6^{\log_{6}n \times \log_{2}6}

     (6^{\log_{6}n})^{\log_{2}6}

     (n)^{\log_{2}6}

     (n)^{\log_{2}(2\times3)}

     (n)^{\log_{2}2 + \log_{2}3}

     (n)^{1+ \log_{2}3}
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  4. #4
    Senior Member vincisonfire's Avatar
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    Quote Originally Posted by Mush View Post
     (n)^{1+ \log_{2}3}
    Certainly a better answer.
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