Question

Express log_4 27 - 2(log_4 3) as a single logarithm.

Answer 1

you know the log properties?

Answer 2

I don't really understand them

Answer 3

I'm thinking it is either log_4 12
or
log_4 18

Answer 4

I'm gonna attach a file

Answer 5

okay!

Answer 6

Ricardo: Do you happen to have a reference of any kind (book, Internet or whatever) that lists the three main properties of logs that apply to this kind of question?
If so, that'd be really helpful.

Answer 7

yes, they are not difficult to use, just take a lookon google

Answer 8

I think I understand it a bit more. I am terrible at this sort of thing.

Answer 9

it's the same as the distributive property, a(b+c) = ab + ac, you just have to follow the rules

Answer 10

Yeah! I see that now! Can you help me with this one?

Solve 3^(2x) = 30.

log_3 (30/2)
log_3 15
2(log_3 30)
log_3 60

Answer 11

R: take the log of both sides. You certainly can, if you wish, take the log to base 3 of both sides, ending up with 2x=log_3 30.

How muich sense does this make to you?

Answer 12

It makes a good amount of sense. So now if I have

2x=log_3 30

I would divide boths sides by 2x and get x = log_3 15 ?

Answer 13

Yes, divide both sides by 2; but the right side ends up as (1/2)log_3 30. Can't divide that 30 by 2. Make a note of this so we can discuss it in more depth when we again have access to Equation Editor and Draw.

Answer 14

Okay, but that answer isn't one of my answer choices that they had given me..

Answer 15

If that's the case, I'd bet your answer choices involve either ln or log, not log_3. Correct?

Answer 16

a. log_3 (30/2)
b. log_3 15
c. 2(log_3 30)
d. log_3 60

Those are the answer choices.

Answer 17

it should be C but 1/2( ) not 2( )

Answer 18

Solve 3^(2x) = 30.
Take log_3 of both sides, resulting in 2x=log_3 30
Solve for x:

[log_3 30]/2 = x. I agree with lucaz.

Answer 19

That's what I was thinking. Maybe the quiz had a glitch or they forgot to put that in.

Answer 20

Ric: Again, I 'd suggest you and I (or you and others) practice application of these and other rules of logs once we again have access to Equation Editor and Draw.

Care to discuss any other practice problems right now?

Answer 21

Nope, All of the rest I know I can do on my own. Thank you to the both of you!

Answer 22

no problem..

Answer 23

You're welcome!