Answer
$\log_2{(1+1)} \ne \log_2{1} + \log_2{1}$
Refer to the step-by-step part for the computations.
Work Step by Step
If $x=1$ and $y=1$, then
$\log_2 {(x+y)}=\log_2{(1+1)}=\log_2{2}=1$
$\log_2{x}+\log_2{y}=\log_2{1}+\log_2{1}=0+0=0$
Thus, in general,
$\log_2(x+y)\ne \log_2{x}+\log_2{y}$.
