`log_a c=b<=>a^b=c `

Z definicji `log_2 (1/4)=−2`, bo `2^(−2)=1/(2^2)=1/4` `log_2 2=1`, bo `2^1=2` Zatem `log_2 (1/4)−log_2 2=−2−1=−3`