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Math Help - logarithms

  1. #1
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    logarithms

    Evaluate \log_{9} 49-\log_{3} 7
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by requal View Post
    Evaluate \log_{9} 49-\log_{3} 7
    Recall these formulae :
    \log_a(b)=\frac{\ln(b)}{\ln(a)}
    \log(a^b)=b \log(a)

    So \log_9 49=\frac{\ln(49)}{\ln(9)}=\frac{\ln(7^2)}{\ln(3^2)  }=\frac{\ln(7)}{\ln(3)}=\log_3(7)

    Thus...
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  3. #3
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    Hello, requal!

    Evaluate: . \log_{9} 49-\log_{3} 7

    \text{Let: }\,\log_949 \:=\:p .[1]

    Then we have: . 9^p \:=\:49 \quad\Rightarrow\quad (3^2)^p \:=\:7^2 \quad\Rightarrow\quad 3^{2p} \:=\:7^2

    Take the square root of both sides: . \left(3^{2p}\right)^{\frac{1}{2}} \:=\:\left(7^2\right)^{\frac{1}{2}} \quad\Rightarrow\quad 3^p \:=\:7

    This can be written: . p \:=\:\log_37 .[2]


    From [1] and [2], we have: . \log_949 \:=\:\log_37

    . . Therefore: . \log_949 - \log37 \;=\;0

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