Question

Помогите: Log27 8/ Log3 4

Answer 1

$\frac{Log_{27}8}{Log_34}= \frac{Log_{3^3}8}{Log_34}= \frac{1}{3}* \frac{Log_38}{Log_34} = \frac{1}{3}*Log_48= \frac{1}{3}Log_{2^2}8= \frac{1}{3}* \frac{1}{2}*Log_28=$

$= \frac{1}{3}* \frac{1}{2}*Log_22^3= \frac{3}{6}= \frac{1}{2}=0.5$