Question 1012071
Hi there. I'm sorry to hear about your situation, let me try my best.
Luckily, there's not a lot you have to know for these problems, but here's what you do have to know.
{{{log(ab)=log(a)+log(b)}}}
{{{log(a/b)=log(a)-log(b)}}}
{{{log(a^b)=b*log(a)}}}
With this, we can tackle the problems.

Six is really just 2 times 3, so {{{log(6)=log(2*3)}}}. We have from our first identity that {{{log(ab)=log(a)+log(b)}}}, so {{{log(2*3)=log(2)+log(3)}}}. We know these values, and we can then use substitution. {{{log(2)+log(3)=0.301+0.447=0.748}}}.


The second one is easier, as 2/3 is really straightforward. We use the identity {{{log(a/b)=log(a)-log(b)}}} here. We plug in a as 2 and b as 3 because they are the numerator and denominator respectively. We get {{{log(2/3)=log(2)-log(3)}}}. Then we can use substitution of values to solve. {{{log(2)-log(3)=0.301-0.447=-0.146}}}.


Don't get tricked up on the third one because it's a decimal, remember that 1.5 is the same thing as 3/2 in fractional form. {{{log(1.5)=log(3/2)}}}. We use the same identity as above. {{{log(3/2)=log(3)-log(2)}}} {{{log(3)-log(2)=0.447-0.301=0.146}}}.


This one's a bit harder, but we just have to go back to the basics. Factorize, and you'll get {{{18=2*3^2}}}. Therefore, {{{log(18)=log(2*3^2)}}}. We can break it apart. We get {{{log(2*3^2)=log(2)+log(3^2)}}}, which due to {{{log(a^b)=b*log(a)}}}, becomes {{{log(2)+log(3^2)=log(2)+2*log(3)}}}. Substitute in your values, and you're done. {{{log(2)+2*log(3)=0.301+2(0.447)=1.195}}}