Question 1089582
given:{{{log(2) = 0.3}}} and	{{{log(3) = 0.48}}} (I guess all is in base {{{10}}})
Find:
1. 
{{{log(12)}}}
={{{log(2^2*3)}}}
={{{log(2^2)+log(3)}}}
={{{2log(2)+log(3)}}}
={{{2*0.3+0.48}}}
={{{0.6+0.48}}}
={{{1.08}}}

2. 

{{{log(1/18)}}}

={{{log(1/(2*3*3))}}}

={{{log(1)-log(2*3*3)}}}
={{{0-(log(2)+log(3)+log(3))}}}
={{{-(0.3+0.48+0.48)}}}
={{{-1.26}}}


3. loga (\sqrt[3]{9})-> if this is {{{log(sqrt(3)/9)}}}, you will have

{{{log(sqrt(3)/9)}}}

={{{log(sqrt(3))-log(9)}}}

={{{log(3^(1/2))-log(3^2)}}}

={{{(1/2)log(3)-2log(3)}}}

={{{(1/2-2)log(3)}}}

={{{-(3/2)log(3)}}}

={{{-(3/2)*0.48}}}

={{{-(1.44/2)}}}

={{{-0.72}}}