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Math Help - solving for k

  1. #1
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    solving for k

    {e^{\frac{{{{\log }_2}k}}<br />
{{10}}}} = 1<br />
    {e^{\frac{{\ln k}}<br />
{{\frac{{\ln 2}}<br />
{{10}}}}}} = 1<br />
    {e^{\frac{{\ln k}}<br />
{{10\ln 2}}}} = 1<br />
    \ln {e^{\frac{{\ln k}}<br />
{{10\ln 2}}}} = \ln 1<br />
    \frac{{\ln k}}<br />
{{10\ln 2}} = \ln 1<br />
    \ln k = 10(\ln 1)(\ln 2)<br />
    {e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}<br />
    k = 1<br />

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

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

    \ln {e^{\frac{{{{\log }_2}k}}<br />
{{10}}}} = \ln 1 so \frac{\log_2 k}<br />
{10} = 0 so 2^{\log_2 k} = 2^0 so k = 1.
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