Condense the sum ${\log }_2\left(5\right)+{\log }_2\left(8\right)$ into a single logarithm.

${\log }_2\left(13\right)$

${\log }_{10}\left(40\right)$

${\log }_2\left(5\right)\cdot {\log }_2\left(8\right)$

${\log }_2\left(40\right)$