1. ## 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?

2. Originally Posted by genlovesmusic09
$\displaystyle {e^{\frac{{{{\log }_2}k}} {{10}}}} = 1$
$\displaystyle {e^{\frac{{\ln k}} {{\frac{{\ln 2}} {{10}}}}}} = 1$
$\displaystyle {e^{\frac{{\ln k}} {{10\ln 2}}}} = 1$
$\displaystyle \ln {e^{\frac{{\ln k}} {{10\ln 2}}}} = \ln 1$
$\displaystyle \frac{{\ln k}} {{10\ln 2}} = \ln 1$
$\displaystyle \ln k = 10(\ln 1)(\ln 2)$
$\displaystyle {e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}$
$\displaystyle k = 1$

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 $\displaystyle k = 1.$