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Math Help - Need help with logarithm equation.

  1. #1
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    Need help with logarithm equation.

    Given that m=2^x and n=2^y , express (log2)(Squareroot (m^3)n) / 64 in terms of x and y.

    So far i got
    1/2 (log2)m^3 n - (log2)2^6

    1/2 (log2)m^3 n - 6

    HELP!~
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help with logarithm equation.

    I guess the expression is:
    \log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)

    What you've done already is fine:
    = \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6

    Now use \log(a\cdot b)=\log(a)+\log(b)
    And afterwards where necessary the definition to simplify:
    \log_a(y)=x \Leftrightarrow a^{x}=y
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  3. #3
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    Re: Need help with logarithm equation.

    Quote Originally Posted by Siron View Post
    I guess the expression is:
    \log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)
    I would interpret it as
    \log_2\left(\frac{n\sqrt{m^2}}{64}\right)
    with m= 2^x and n= 2^y that is
    \log_2\left(\frac{2^{3x/2}2^y}{2^6}\right)
    \log_2(2^{3x/2})+ \log_2(2^y)- log_2(2^6)

    What you've done already is fine:
    = \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6

    Now use \log(a\cdot b)=\log(a)+\log(b)
    And afterwards where necessary the definition to simplify:
    \log_a(y)=x \Leftrightarrow a^{x}=y
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  4. #4
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    Re: Need help with logarithm equation.

    yes , the expression is

    (log2) 2^3/2x + (log2)2^1/2y - 6

    so the answer is (3/2)x + (1/2)y -6
    Thanks for helping !
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Need help with logarithm equation.

    You're welcome and enjoy the forum!
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