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Thread: solving for k

  1. #1
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    solving for k

    $\displaystyle {e^{\frac{{{{\log }_2}k}}
    {{10}}}} = 1
    $
    $\displaystyle {e^{\frac{{\ln k}}
    {{\frac{{\ln 2}}
    {{10}}}}}} = 1
    $
    $\displaystyle {e^{\frac{{\ln k}}
    {{10\ln 2}}}} = 1
    $
    $\displaystyle \ln {e^{\frac{{\ln k}}
    {{10\ln 2}}}} = \ln 1
    $
    $\displaystyle \frac{{\ln k}}
    {{10\ln 2}} = \ln 1
    $
    $\displaystyle \ln k = 10(\ln 1)(\ln 2)
    $
    $\displaystyle {e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}
    $
    $\displaystyle k = 1
    $

    is this correct?
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    $\displaystyle {e^{\frac{{{{\log }_2}k}}
    {{10}}}} = 1
    $
    $\displaystyle {e^{\frac{{\ln k}}
    {{\frac{{\ln 2}}
    {{10}}}}}} = 1
    $
    $\displaystyle {e^{\frac{{\ln k}}
    {{10\ln 2}}}} = 1
    $
    $\displaystyle \ln {e^{\frac{{\ln k}}
    {{10\ln 2}}}} = \ln 1
    $
    $\displaystyle \frac{{\ln k}}
    {{10\ln 2}} = \ln 1
    $
    $\displaystyle \ln k = 10(\ln 1)(\ln 2)
    $
    $\displaystyle {e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}
    $
    $\displaystyle k = 1
    $

    is this correct?
    Yes but why switch to the natural log.

    $\displaystyle \ln {e^{\frac{{{{\log }_2}k}}
    {{10}}}} = \ln 1 $ so $\displaystyle \frac{\log_2 k}
    {10} = 0$ so $\displaystyle 2^{\log_2 k} = 2^0 $ so $\displaystyle k = 1.$
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