Question

log2(27)-2 log2(3)+log2(2/3)

Answer 1

$\log_227-2 log_23+log_2frac{2}{3}=\ log_227- log_23^2+log_2frac{2}{3}=\ log_227- log_29+log_2frac{2}{3}=\ log_2frac{27}{9}+log_2frac{2}{3}=\ log_23+log_2frac{2 }{3}=\ log_2(3cdot frac{2 }{3})=\ log_22=\1$

Answer 2

log₂(27)- 2 log₂(3)+log₂ (2/3) = log₂(27)- log₂(3²)+log₂ (2/3) = = log₂(27)- log₂(9) +log₂ (2/3) = log₂(27 : 9 *  2/3) = log₂(3 * 2/3) = log₂(2) = 1