Question

logb(sqrt(27))-logb(sqrt(32))

Answer 1

what is the goal here? are you supposed to write this as a single log?

Answer 2

I'll attach question

Answer 3

or are you supposed to expand it even further? for example
\[\sqrt{27}=3\sqrt3\] and
\[\sqrt{32}=4\sqrt2\]

Answer 4

yeah that would be good, see the actual question

Answer 5

well that is a whole different story

Answer 6

since \[\sqrt{27}=3\sqrt3\] we have
\[\log(\sqrt{27})=\log(3\sqrt3)=\log(3)+\log(\sqrt3)=\log(3)+\frac{1}{2}\log(3)\]

Answer 7

since you are told that \(\log(3)=C\) then this translates to
\[c+\frac{1}{2}C=\frac{3}{2}C\]

Answer 8

similar idea for
\[\log(\sqrt{32}=\log(4)+\frac{1}{2}\log(2)\]

Answer 9

oh damn i made a mistake, it is \(\log(3)=A\) not \(C\)
no matter, change the first thing to
\[\frac{3}{2}A\]

Answer 10

then the second part is
\[\log(4)+\frac{1}{2}\log(2)=2\log(2)+\frac{1}{2}\log(2)=2C+\frac{1}{2}C=\frac{5}{2}C\]

Answer 11

that was exhausting
final answer
\[\frac{3}{2}A-\frac{5}{2}C\]

Answer 12

thank you so much. can i ask you about another problem too? because wolfram and i got the same answer but its wrong in webworks so i dont understand