$\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?