Question

name that value

Answer 1

of what?

Answer 2

what grade are you in?

Answer 3

9th

Answer 4

You have to expand the expression you have. Do you know how to expand
\[\log_{b}(\frac{ A^{3}D^{4} }{ C^{2} })\]?

Answer 5

I'm guessing 16875 / 4. It's just a guess.

Answer 6

no could you show me?

Answer 7

Okay. Well, when you have groups of multiplication and division inside of a logarithm, they can be split apart. For 2 things being multiplied inside of a logarithm, it can be split into the addition of those 2 things being multiplied inside separate logs. So for example, if I have:
\[\log_{b}(xy) = \log_{b}x + \log_{b}y\]So x and y were multiplied inside of the log, then it was split into the addition of two logs, each multiply inside their own logarithm. The logarithm of each keeps its base.

A similar idea works with division. If you have 2 items being divided inside of a logarithm, it can be split into a subtraction of those two things inside of a logarithm. So like this:
\[\log_{b}(\frac{x}{y}) = \log_{b}x - \log_{b}y\]

The last thing we need to be able to do is remove exponents. If the ENTIRE expression inside of the logarithm is raised to a power, that power can be moved into the front of the logarithm. So for example:
\[\log_{b}(x+1)^{2} = 2\log_{b}(x+1)\] But again, that power must be applied to the whole expression inside, or else you can't rip it out.

Answer 8

Kind of make sense so far?

Answer 9

yea

Answer 10

So think you could attempt to split apart your logarithm expression?

Answer 11

but how do I expand? I'm so lost

Answer 12

You would be trying to use the above 3 rules. Multiplication means split apart with addition of logs and division means split apart with subtraction of logs. Like Ill do one of the steps for you so you can see.

So, I have that C^2 on bottom, so division. That means I can put C^2 in a separate logarithm and have it be subtracted. So that means I can do:
\[\log_{b}(\frac{ A^{3}D^{4} }{ C^{2} }) = \log_{b}(A^{3}D^{4}) - \log_{b}(C^{2})\]

Can you see what I did and why I was able to do it?

Answer 13

My brain is so scrambled I'm sorry I'm getting confused

Answer 14

Do any of the rules I posted above make sense, or is everything Ive said confusing?

Answer 15

it's kind of confusing

Answer 16

Well, before you can really even do this problem, you'll have to be able to understand each rule. This problem makes you use 3 rules, which may be too much if none of this makes sense. So if I look at the rule with multiplication, where it becomes addition:
\[\log_{b}(x^{2}y) = \log_{b}(x^{2})+\log_{b}(y)\]
\[\log_{b}(2*3) = \log_{b}(2) + \log_{b}(3)\]

Bleh, gotta go, I apologize. Maybe someone can pick up from here. Good luck ^_^

Answer 17

thanks.