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Math Help - Solving equation with logs

  1. #1
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    Solving equation with logs

    Solving equation with logs-screen-shot-2010-09-12-2.02.05-pm.png

    With the help of technology I've figured out that r=27 but I can't quite work it out.

    I know 2^3log4(3)=2^log4(27 or 2^(log27)/(log4) and sqrt(r)=r^(1/2) but I can't quite put it all together....
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  2. #2
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    First notice that \log_4(2)=\frac{1}{2}
    Then  \displaystyle 2^{3\log_4(3)}= 2^{\log_4(27)} .
    Take the log of both \log_4(27)\log_4(2)= \frac{1}{2}\log_4(27)= \log_4(\sqrt{27})=\log_4(\sqrt{r})
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  3. #3
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    Hello, metaname90!

    \text{Solve for }r\!:\;\;2^{3\log_4(3)} \:=\:\sqrt{r}

    The left side is: . 2^{\log_4(3^3)} \;=\;2^{\log_4(27)}


    Since 2 = 4^{\frac{1}{2}}
    . . we have: . (4^{\frac{1}{2}})^{\log_4(27)} \;=\; 4^{\frac{1}{2}\log_4(27)} \;=\;4^{\log_4(27^{\frac{1}{2}})} \;=\;27^{\frac{1}{2}}

    The equation becomes: . 27^{\frac{1}{2}} \:=\:r^{\frac{1}{2}} \quad\Rightarrow\quad r \:=\:27
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