Question

(31g2+1g0,25):(1g14-1g7)

Answer 1

$\frac{3\, lg\, 2+lg\, 0,25}{lg\, 14-lg\, 7}=\frac{lg\, (8\cdot 0,25)}{lg\, (14:7)}=\frac{lg2}{lg2}=1$