Question

Can someone please help me simplify this logarithm:

4log(3) 1/3 + 2log(27) 9

1/3 and 9 are the arguments.

Answer 1

hint:
\[1) \; 4\log_3 (\frac 13) \implies 4\log_3 (3^{-1}) \implies 4(-1) \implies -4\]
\[2) \; 2 \log_{27} (9) \implies 2\log_{27} (27^{2/3}) \implies 2 (\frac{2}{3}) \implies \frac 43\]
does that help?

Answer 2

Yes thank you you are amazing . But how do I make both of the bases equal??

Answer 3

which base?

Answer 4

do you mean what i did with \[2\log_{27} (27^{2/3})\]

Answer 5

you'd have to use change of base formula to do that.....
log(b) A = log A/ log b

Answer 6

im confused with this question....is this evaluating the logarithm? or is it condensing?

Answer 7

Evaluating. It has an equal sign at the end and a "?"

Answer 8

oh. then what do you mean by how to make the bases same?

Answer 9

Well to solve this further and apply properties etc, don't I have to establish a common base?

Answer 10

Lol, this is my summer assignment for Pre Calc and I completely forgot all the Algebra II.

Answer 11

well...unless your condensing...you dont need to make the bases the same. you can just do what i did

Answer 12

maybe I'm condensing I'm not sure. I just want to make this expression as simplified as possible and after what you did (thankyou) I got a step further but I don't know wht to do next.

Answer 13

but...if you insist...
\[\large 4\log_3 (\frac 13) \implies 4 \frac{\log_{27} (\frac 13)}{\log_{27} (3)} \implies 4 \frac{\log_{27} (3^{-1})}{\log_{27} (27^{1/3})} \implies 4 \frac{\log_{27} (27^{-1/3})}{\log_{27} (27^{1/3})}\]
\[\large \implies 4(\frac{-1/3}{1/3} \implies 4(-1) \implies 4\]
just a lengthy way to solve lol

Answer 14

anyway.... what you do now is \[-4 + \frac 43\]
then you have the answer

Answer 15

Your phenomenal. I don't get it, but your fantastic. Thnk you for your time!! I will try to go over it. But you know, if you want to continue helping me, can you explain your steps? Hehe :P

Answer 16

lol nevermind that second solution i did...just focus on the first one

Answer 17

my first comment is the simplest way

Answer 18

anyway i have to go now sorry

Answer 19

Alright. I get it. Thanks <3 !!!

Answer 20

welcome