Question

Vreau și eu rezolvarea va rog mult ‼️‼️‼️

Answer 1

$\it \dfrac{log_3 2+3\cdot log_3 0,25}{log_328-log_3 7}=\dfrac{log_3 2+3\cdot log_3\dfrac{1}{4}}{log_3 \dfrac{28}{7}}=\dfrac{log_3 2+log_3\Big(\dfrac{1}{4}\Big)^3}{log_3 4}=\\ \\ \\ =\dfrac{log_3 2\cdot\dfrac{1}{64}}{log_3 4}=\dfrac{log_3 \dfrac{1}{32}}{log_3 4}=\dfrac{log_3 2^{-5}}{log_3 2^2}=\dfrac{-5\cdot log_3 2}{2\cdot log_3 2}=-\dfrac{5}{2}=-2,5$