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Thread: Simplifying and/or Computing problems

  1. #1
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    Simplifying and/or Computing problems

    Hi I need help simplifing these problems.

    1. 2ln$\displaystyle sqrt(e)$
    = $\displaystyle sqrt(3)^2$
    = 3

    2$\displaystyle ^Logx$ = 2^log4 x
    = 4$\displaystyle ^3x$

    3. tan$\displaystyle ^-1$(1/$\displaystyle sqrt(3)$
    = pi/3

    are these right?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by killasnake View Post
    1. 2ln$\displaystyle sqrt(e)$
    = $\displaystyle sqrt(3)^2$
    = 3
    Where did the 3 come from?

    $\displaystyle 2~ln(\sqrt{e}) = 2~ln \left ( e^{1/2} \right ) = 2~ \frac{1}{2} \cdot ln(e) = 1 \cdot 1 = 1$

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by killasnake View Post
    2$\displaystyle ^Logx$ = 2^log4 x
    = 4$\displaystyle ^3x$
    I presume "Log" is $\displaystyle log_{10}$?

    Use the change of base formula:
    $\displaystyle log_{10}(x) = \frac{log_2(x)}{log_2(10)}$

    So
    $\displaystyle 2^{log_{10}(x)} = 2^{\frac{log_2(x)}{log_2(10)}}$

    $\displaystyle = \left ( 2^{log_2(x)} \right )^{1/
    log_2(10)}$

    $\displaystyle = x^{1/log_2(10)} \approx x^{0.30103}$

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by killasnake View Post
    3. tan$\displaystyle ^-1$(1/$\displaystyle sqrt(3)$
    = pi/3
    Close.
    $\displaystyle tan^{-1} \left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi}{6}$

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Where did the 3 come from?

    $\displaystyle 2~ln(\sqrt{e}) = 2~ln \left ( e^{1/2} \right ) = 2~ \frac{1}{2} \cdot ln(e) = 1 \cdot 1 = 1$

    -Dan
    oh crap i just realised my problem... the 3 was suppose to be an e
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