Question

log2bază28-2log2baza14+log2bază7

Answer 1

$\displaystyle log_228-2log_214+log_27=log_228-log_214^2+log_27= \\ =log_2 \frac{28}{14^2}+log_27=log_2 \frac{2^2 \cdot 7}{2^2 \cdot 7^2} +log_27=log_2 \frac{1}{7} +log_27= \\ =log_27 \cdot \frac{1}{7} =log_2 \frac{7}{7} =log_21=0$