Question

1) log3 18 + log3 (1\18) 2) 2log2 1\4 - 3log1\3 27

Answer 1

1)

$log_{3}18+log_{3}\frac{1}{18}=log_{3}(18\cdot \frac{1}{18}) =log_{3}1=0$

2)

$2log_{2}\frac{1}{4} -3log_{\frac{1}{3} }27=2log_{2}2^{-2} -3log_{\frac{1}{3} }(\frac{1}{3})^{-3} =2\cdot (-2)log_{2}2-3\cdot (-3)log_{\frac{1}{3} }\frac{1}{3}=\\\\=-4+9=5$