Question

Logarithm... write with same base

Answer 1

\[\log _{2}12+\log _{4}6-\log _{2}24\]

Answer 2

So obviously I start with the middle log... that has the base of 4 to change it to base 2...

\[4^{x}=6\]
\[(2^{2})^{x}=6\]
\[2^{2x}=6\]

Answer 3

Which means...

\[\log _{2}6=2x\]

Answer 4

How do I simplify this even more?

Answer 5

I know the answer is... \[\frac{ 1 }{ 2 }\log _{2}6 = \log _{2}6^{\frac{ 1 }{ 2 }}\]

Answer 6

I just don't know how you jump to that...

Answer 7

@MathLegend
Try to combine what you can for now
Then simplify it

Answer 8

use the division property first
log(a)-log(b)=log(a/b)
for the ones that have the same base

Answer 9

ok

Answer 10

\[\log _{2}(\frac{ 12 }{ 24 })+\log _{4}6\]

Answer 11

great so far
but 12/24 reduces

Answer 12

Oh, I wasn't sure if we were allowed to reduce.. because my teacher never does with logs.

\[\log _{2}(\frac{ 1 }{ 2 })+\log _{4}6\]

Answer 13

ok
so that 1/2 is very important here
because you know log_2(1)=?
and you know log_2(2)=?

Answer 14

\[\log_2(\frac{1}{2})=\log_2(1)-\log_2(2)=? \]

Answer 15

-1 ?

Answer 16

so you have
\[-1+\log_4(6) \]

Answer 17

now if you want to write this as one log with one base
you can

Answer 18

\[1=\log_a(a)\]

Answer 19

where a>0 and not equal to 1 of course :)

Answer 20

Okay, so should I do what I did earlier to have this change to...

\[2^{2x}=6 = \log _{2}6=2x\]

Answer 21

\[\log _{2}6=2x\]

Answer 22

You don't need that

do you mean 2^(2x)=6 implies log_2(6)=2x

Answer 23

replace the 1 in your problem with log_4(4)
so the bases will match

Answer 24

\[-1+\log_4(6)=-\log_4(4)+\log_4(6)=\log_4(6)-\log_4(4)\]
you should be able to take it from here

Answer 25

\[\log _{4}(\frac{ 3 }{ 2 })\]

Answer 26

that's great

Answer 27

But the answer says...

\[\log _{2}(\frac{ \sqrt{6} }{ 2 })\]

Answer 28

it looks like they also wrote it with base 2 instead of base 4
so if you want it in base 2 you will need to play with our answer some more

Answer 29

\[\log_4(\frac{3}{2})=\frac{\log_2(\frac{3}{2})}{\log_2(4)} \text{ (By change of base formula ) }\]
\[=\frac{\log_2 (\frac{3}{2})}{\log_2(2^2)}\]
do you got from here?

Answer 30

I'm confused on how you just went from base 4 to base 2 in the numerator and the denominator... because isn't the formula...

\[\log _{b}M=\frac{ logM }{ logb }\]

Answer 31

\[\frac{ \log4 }{ \log \frac{ 3 }{ 2 } }\]

Answer 32

Am I utilizing the formula incorrectly?

Answer 33

Ok sorry was eating...

but yeah the formula is
\[\log_b(M)=\frac{\log_a(M)}{\log_a(b)}\]
We want a to be 2 so I replaced a's with 2
M was 3/2 to I replaced M with 3/2 everywhere else
b was 4 so I replaced all the b's with 4

Answer 34

when comparing these two
you should see that..