Question

Question on how to simplify all the logs have a base 12
so log9+log16x-2logx

Answer 1

\(\Large\color{midnightblue}{ \rm log_{12}9+log_{12}16x-2~log_{12}x }\)
\[\log_ab+\log_ac=\log_a(b \times c)~~~~and\]\[\log_ab - \log_ac=\log_a(b \div c)\]

Answer 2

the black equations are the rules.

Answer 3

ok got it its equal to 9*16x/x^2?

Answer 4

Yes, \(\Large\color{midnightblue}{ \rm log_{12}(9 \times 16x \div x ) }\)

Answer 5

Your answer will be a whole number, just simplify it.

Answer 6

wait how is it a whole number?

Answer 7

\(\Large\color{darkgreen}{ \rm log_{12}(144) = }\)
\(\Large\color{darkgreen}{ \rm 2~log_{12}(12) = }\)
\(\Large\color{darkgreen}{ \rm 2 .}\)

Answer 8

I thought that was it and isn't it \[x ^{2}\] on the bottom?

Answer 9

It is, actually... so it should boil down to
\[\Large \log_{12}\left(\frac{144}{x}\right)\]
You could always separate them again, if that looks more aesthetically pleasing to you...
\[\Large = \log_{12}(144) - \log_{12}(x)\]

Answer 10

I thought, the `x`s cancel out.

\(\Large\color{darkgreen}{ \rm log_{12}(9 \times 16x \div x) }\)

Answer 11

They do indeed.
But you forget that that's a \(\large ... - \color{red}2\log_{12}(x) \) there at the rightmost end...

Answer 12

so its 2-log(x)

Answer 13

That is correct, @eragon4 ^_^

Answer 14

yes.. I forgot that -:(

Answer 15

than solve for x right? or its says simplify so do I leave it that way?

Answer 16

You can't solve for x unless you have an equation ^^

Answer 17

Is all this... equal to something?

Answer 18

no I thought I cold make it -logx=-2 and solve it

Answer 19

haha... no :)

You won't be able to solve anything without an equation to begin with :D

Answer 20

err... I meant solve FOR anything lol

Answer 21

whoops ok thanks

Answer 22

No problem.
Sleepy now....
Signing off...
Best of luck :)
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Terence out