If log72 = m, then log4928 is equal to ?

$\log 4928 = \frac{1}{2\log 77\times 4}$

= $\frac{1}{2}+\frac{1}{2}2\log 72$
= $\frac{1}{2}+\log 72$
=  $\frac{1}{2} + m$
=  $\frac{1+2m}{2}$.