Question

\(\large \frac{log24\sqrt{3}- \frac{3}{2}}{log{12}-1} \) simplify

Answer 1

\[\huge \frac{log24\sqrt{3}- \frac{3}{2}}{log{12}-1} \]

Answer 2

do we know the base of the log?

Answer 3

10

Answer 4

\[\frac{\log(24)\sqrt3-\frac32}{\log(12)-1}\]radical 3 is outside the root, right?

Answer 5

outside the log I mean

Answer 6

or is it inside the log?

Answer 7

it's inside the log

Answer 8

lemme type out the whole question.... i'm probably taking the wrong path

Answer 9

I think I got it...

Answer 10

but when i type the whole question and where i've gotten to into the calculator, i get the same answer...

Answer 11

\[\frac{\log(24\sqrt3)-\frac32}{\log(12)-1}=\frac{\log(2^3\cdot3^{3/2})-\frac32}{\log(2^2\cdot3)-1}\]is this at all like what you did?

Answer 12

\[=\frac{\log(2^3)+\log(3^{3/2})-\frac32}{\log(2^2)+\log(3)-1}\]\[=\frac{3\log2+\frac32\log3-\frac32}{2\log2+\log3-1}\]\[=\frac{6\log2+3\log3-3}{4\log2-2\log3-2}\]not sure what more you want to do to it

Answer 13

just right click and you can add whatever you forgot

Answer 14

too late...

Answer 15

\[\huge \frac{log\sqrt{27}+log8-log\sqrt{1000}}{log{1.2}} \]

Answer 16

I would try to break it all down as much as possible first, looking for powers to pull out of the logs

Answer 17

\[\frac{\log(3^{3/2})+\log(2^3)-\log(10^{3/2})}{\log(\frac65)}\]now apply the log rules

Answer 18

i'll let u know what i get :) thanks

Answer 19

welcome :D

Answer 20

actually the answer is kinda neat, but you have to break it /all/ the way down

Answer 21

\[\frac{\log(3^{3/2})+\log(2^3)-\log((5\cdot2)^{3/2})}{\log(\frac{3\cdot2}5)}\]/now/ apply the log rules

Answer 22

brb

Answer 23

the calculator says the answer is 3/2 but i'm still not sure which rules to apply to get there... i think i'm only applying some of them and i have limited knowledge on the rest so i'm probably not seeing it.... can u help me go down further the solution, or even rather give me the rules i'm supposed to use... :/

Answer 24

sure, you only need three rules\[\log(ab)=\log a+\log b\]\[\log(\frac ab)=\log a-\log b\]\[\log(a^b)=b\log a\]I'll work it, you tell me when to stop...

Answer 25

\[\frac{\log(3^{3/2})+\log(2^3)-\log((5\cdot2)^{3/2})}{\log(\frac{3\cdot2}5)}\]\[\frac{\frac32\log3+3\log2-\frac32\log(5\cdot2)}{\log(3\cdot2)-\log5}\]

Answer 26

multiply by 2/2 and split up the multiples with the first log rule above\[\frac{3\log3+6\log2-3\log(5\cdot2)}{2\log(3\cdot2)-2\log5}\]\[\frac{3\log3+6\log2-3\log5-3\log2}{2\log3+2\log2-2\log5}\]\[\frac{3\log3+3\log2-3\log5}{2\log3+2\log2-2\log5}\]clear now?

Answer 27

why aren't I allowed to punch it into the calculator?

Answer 28

you are, but calculators don't simplify for you...

Answer 29

well the idea of multiplying through never occurred to me... thanks

Answer 30

i meant multiplying through by 2/2

Answer 31

sure thing
In general it is nice to get rid of fractions in the numerator and denominator
once I did that, out popped the answer

Answer 32

:) thanks a ton