Question

@DanJS Need help!

Answer 1

do you remember the log properties?

Answer 2

You the log^M + log^N = log^MN thing? Yeah, I do

Answer 3

\[\log _{b}(x*y) = \log _{b}x + \log _{b}y\]

Answer 4

\[\log _{b}(\frac{ x }{ y }) = \log _{b}x - \log _{b}y\]

Answer 5

\[\log _{b}x^a = a*\log _{b}x\]

Answer 6

\[y = \log _{b}x~~~~~~or~~~~~~b^y=x\]

Answer 7

I know them already, how do I use them

Answer 8

so if you have
\[Log _{b}(\frac{ A^5*C^2 }{ D^6 })\]

How would you rewrite that , using the rules

Answer 9

the answer's 1.2 right?

Answer 10

i dont know, havent calculated it

Answer 11

You can rewrite the thing like this

Answer 12

Nvm, it's not 1.2. I did 3 times 2 = 6 and then added 5 and 2....
Eventually I got 6^7/5^6

Answer 13

\[Log _{b}(\frac{ A^5*C^2 }{ D^6 })=\log _{b}A^5 +\log _{b}C^2 - \log _{b}D^5\]

addition from the multiply rule, and subtraction from the divide rule

Answer 14

now the power rule applies to each term

Answer 15

\[Log _{b}(\frac{ A^5*C^2 }{ D^6 })=\log _{b}A^5 +\log _{b}C^2 - \log _{b}D^5 = 5*\log _{b}A + 2*\log _{b}C-5*\log _{b}D\]

Answer 16

what's the overall answer?

Answer 17

So you have
\[5*\log _{b}A + 2*\log _{b}C-5*\log _{b}D \]

and they give you what each of the log base b of A C and D are

Answer 18

I hope you understand what i did and not just the answer

Answer 19

I'm following..

Answer 20

Plugging in the values they gave you , you get
5*3 + 2*2 - 5*5

Answer 21

\[5*\log _{b}A + 2*\log _{b}C-5*\log _{b}D = 5*3 + 2*2 - 5*5\]

Answer 22

They gave you the
\[\log _{b}A = 3~~~and~~~\log _{b}C=2~~~~~and~~~~~\log _{b}D=5\]

Answer 23

That is as close as i can get to the answer without straight telling you "the answer is". lol
5*3 + 2*2 - 5*5 =

Answer 24

k, welcome, goodluck

Answer 25

did you get A?

Answer 26

@DanJs wrote 5*5 but he meant 6*5
log D^6 = 6Log D = 6*5
follow the steps given