Question

Question in comments..
about logs.
let :

Answer 1

\[\large \log_b A =3 ; \log_b C =2 ; \log_b D = 5 \]
What is
\[\large \log_b \frac{ A^3 D^4 }{ C^2 }\]
I just plug in the value right?

Answer 2

\[\large \log_b \frac{ 3^3 5^4 }{ 2^2 }\]
\[\large \log_b =\frac{ 9 \times 625 }{ 4 } = 1406.25\]

Answer 3

Hardly :P

Answer 4

Oh, what do I do then?

Answer 5

A few properties of logs....
\[\Large \log_b(MN)=\log_b(M)+\log_b(N)\]\[\Large \log_b\left(\frac{M}{N}\right)=\log_b(M) -\log_b(N) \qquad N\ne 0\]\[\Large \log_bM^\color{green}p =\color{green}p\log_b(M)\]

Answer 6

Use these to your advantage, and expand \[\Large \log_b \frac{A^3D^4}{C^2}\]
as best you can, and show me what you get :)

Answer 7

Rule of thumb, so you can do it more quickly (when you get the hang of it, of course)
If it's in the numerator, add its log, if it's in the denominator, subtract its log...

^That doesn't really make sense from a strict viewpoint, but hey, it worked for me :>

Answer 8

So..
\[\large (\log_b A^3 - \log_b D^4) - \log_b C^2\]

Answer 9

oops that should be a plus sign

Answer 10

in the parantheses.

Answer 11

Okay then ^_^
\[\large (\log_b A^3 \color{red}+\log_b D^4) - \log_b C^2\]

Answer 12

Now, use that third property, the one involving exponents in logs...

Answer 13

\[\Large \log_bM^\color{green}p =\color{green}p\log_b(M)\]

Answer 14

\[\large 3 \log_b A + 4 \log_b D - 2 \log_b C\]

Answer 15

Yup... NOW, I believe we have specific known value for each of these expressions...

\[\large 3 \color{red}{\log_b A }+ 4 \color{blue}{\log_b D }- 2 \color{green}{\log_b C}\]
substitute away ^_^

Answer 16

Oh!
\[\large 3(3) + 4(5) - 2(2) = 29 - 4 = 25\]

Answer 17

Thank you, that was a big help in understanding this :D

Answer 18

...bingo :)

Answer 19

Your first answer was way off, don't you think? XD

Answer 20

Haha it was lol