1. ## solving for k

${e^{\frac{{{{\log }_2}k}}
{{10}}}} = 1
$

${e^{\frac{{\ln k}}
{{\frac{{\ln 2}}
{{10}}}}}} = 1
$

${e^{\frac{{\ln k}}
{{10\ln 2}}}} = 1
$

$\ln {e^{\frac{{\ln k}}
{{10\ln 2}}}} = \ln 1
$

$\frac{{\ln k}}
{{10\ln 2}} = \ln 1
$

$\ln k = 10(\ln 1)(\ln 2)
$

${e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}
$

$k = 1
$

is this correct?

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

${e^{\frac{{\ln k}}
{{\frac{{\ln 2}}
{{10}}}}}} = 1
$

${e^{\frac{{\ln k}}
{{10\ln 2}}}} = 1
$

$\ln {e^{\frac{{\ln k}}
{{10\ln 2}}}} = \ln 1
$

$\frac{{\ln k}}
{{10\ln 2}} = \ln 1
$

$\ln k = 10(\ln 1)(\ln 2)
$

${e^{\ln k}} = {e^{10(\ln 1)(\ln 2)}}
$

$k = 1
$

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

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