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Math Help - Help with basic Logarithms

  1. #1
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    Help with basic Logarithms

    Just a side question: Why is Log9(3) = 1/2? Can you please solve it step by step for me. Thanks

    1) Solve showing the steps.

    2) Solve showing all steps.

    Thankyou!
    Last edited by BG5965; November 15th 2008 at 03:38 AM.
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  2. #2
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    Quote Originally Posted by BG5965 View Post
    Just a side question: Why is Log9(3) = 1/2? Can you please solve it step by step for me. Thanks

    1) Solve showing the steps.

    2) Solve showing all steps.

    Thankyou!
    To your first question: You are looking for an exponent which belongs to the base 9 such that the result is 3:

    9^x = 3~\implies~9^x = \sqrt{9} ~\implies~9^x = 9^{\frac12}

    2 powers with the same base are equal if the exponents are equal too. Therefore

    x = \dfrac12

    to #1:

    \dfrac{\log_7(32)}{\log_7(8\sqrt{2}}=\dfrac{5\log_  7(2)}{3.5\log_7(2)}= \dfrac{10}7

    to #2:

    \dfrac{\log_a(18)+\log_a(12)-\log_a(125)}{\log_a(42) - \log_a(35)} = \dfrac{2\log_a(3)+\log_a(2)+2\log_a(2)+\log_a(3)-3\log_a(5)}{\log_a(6)+ \log_a(7) - \log_a(7)-\log_a(5)} = = \dfrac{3\left(\log_a(2)+\log_a(3)-\log_a(5)  \right)}{2\log_a(7)+\log_a(6)-\log_a(5)  }
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  3. #3
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    Hello, BG5965!

    2)\;\;\frac{\log_a(18) + \log_a(12) - \log_a(125)}{\log_a(42) - \log_a(35)}

    The numerator is: . \log_a(2\cdot3^2) + \log_a(2^2\cdot3) - \log_a(5^3)

    . . = \;\log_a(2) + \log_a(3^2) + \log_a(2^2) + \log_a(3) - \log_a(5^3)

    . . = \;\log_a(2) + 2\log_a(3) + 2\log_a(2) + \log_a(3) - 3\log_a(5)

    . . = \;3\log_a(2) + 3\log_a(3) - 3\log_a(5)

    . . = \;3\bigg[\log_a(2) + \log_a(3) - \log_a(5)\bigg]


    The denominator is: . \log_a(2\cdot3\cdot7) - \log_a(5\cdot7)

    . . = \;\bigg[\log_a(2) + \log_a(3) + \log_a(7)\bigg] - \bigg[\log_a(5) + \log_a(7)\bigg]

    . . = \;\log_a(2) + \log_a(3) - \log_a(5)


    The fraction becomes: . \frac{3\left[\log_a(2) + \log_a(3) - \log_a(5)\right]} {\log_a(2) + \log_a(3) - \log_a(5)} \;=\;\boxed{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    An alternate method . . .

    We have . \frac{\log_a(18) + \log_a(12) - \log_a(15)}{\log_a(42) - \log_a(35)} \;= \;\frac{\log_a\left(\dfrac{18\cdot12}{125}\right)}  {\log_a\left(\dfrac{42}{35}\right)}

    . . . . . . = \;\dfrac{\log_a(1.728)}{\log_a(1.2)} \;=\;\log_{1.2}(1.728)


    Since 1.2^3 \:=\:1.728, then: . \log_{1.2}(1.728) \;=\;3

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