Question

help me to find the answers for the logarithms guys​

Answer 1

Answer:

$log_{3}(1 \div9 ) = log_{3}(1 ) - log_{3}(9)$

as log of 1 to any base is 0

$log_{3}(1 \div 9) = - log_{3}( {3}^{2} )$

by using property of log

$log_{3}( {3 }^{2} ) = 2 log_{3}(3)$

$log_{3}(3) = 1$

$log_{3}(1 \div 9) = - 2$