Question

are the steps for this equation correct? equation follows:
2 log_2 2 + 2 log_2 6 - log_2 3x = 3
log_2 2^2 + log_2 6^2- log_2 3x = 3
log_2 ((2^2*6^2)/3x)=3
log_2 ((4*36)/3x)=3
log_2 (144/3x)=3
but I get stuck after this. And I'm not even sure if this is correct. someone please help? :)

Answer 1

just to check, this is your work?
\[2 log_2 2 + 2 log_2 6 - log_2 3x = 3
log_2 2^2 + log_2 6^2- log_2 3x = 3
\]\[log_2 ((2^2*6^2)/3x)=3
log_2 ((4*36)/3x)=3
log_2 (144/3x)=3\]

Answer 2

yeah thats it :)

Answer 3

ok so, why in the second line, is your 2and 6 squared?

Answer 4

then why do you have a 3 as a coefficient?

Answer 5

because i was following the third rule of logarithmic functions
http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif

Answer 6

good, (I want you to prove your steps to me just like that when I ask these sort of weird questions. I swear there is a point to it)

Answer 7

because that is how the equation was written

Answer 8

haha alright

Answer 9

? wait, what is the original? i thought it was just \[2 log_2 2 + 2 log_2 6 - log_2 3x = 3

\]

Answer 10

oops, nvm saw the 3 from the previous step

Answer 11

oh haha ok

Answer 12

so that first step is correct now can you justify \[2 log_2 2 + 2 log_2 6 - log_2 3x = 3~~~
log_2 2^2 + log_2 6^2- log_2 3x = 3~~~
\color{red}{log_2 ((2^2*6^2)/3x)=3}
\]

Answer 13

\[2 log_2 2 + 2 log_2 6 - log_2 3x = 3~~~\]
\[log_2 2^2 + log_2 6^2- log_2 3x = 3~~~\]
\[\color{red}{log_2 ((2^2*6^2)/3x)=3}\]

Answer 14

(I trust you can do the basic algebra involved, so just reference the rules and a brief exp)

Answer 15

right. i got stuck on the fact that i have to solve for x in this equation. i simplified it down to \[\log_{2}\left(\begin{matrix}144/3x\end{matrix}\right)=3\] but i dont know what the next step would be to get x by itself. do i get rid of the log or simplify the parenthesis (which confuses me)

Answer 16

well if 144 is whole when you divide by 3 then you can simplify, but we can now get rid of that log

Answer 17

so if you had \(e^{ln(x)}=?\) could you simplify that?

Answer 18

no....lol that looks like complete gibberish. but if i have \[144\div3x=3\] i could multiply both sides by 3x and have \[144=9x\] and simplify it from thee to be \[x=16\] is this correct?

Answer 19

no can't do that

Answer 20

logs are functions

Answer 21

you can't just pull out the parentheses

Answer 22

ie it could just be \(log_2 x=3\) solve for x

Answer 23

it's a one step solution, so anywho watch this: https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_properties/v/introduction-to-logarithm-properties

Answer 24

then check out the inverse of a log

Answer 25

but what happens to the 144/3x if you turn it into \[\log_{2} x=3\]????

Answer 26

not yet, that was a counterex

Answer 27

just an easier ex to get the basic concept accross

Answer 28

oh!!!! i figured it out! thank you :D

Answer 29

k show me your work so i can make sure you get it

Answer 30

x=6 because you change the log formula to an exponential formula so that it reads\[2^{3}=144\div3x\] then multiply both sides by 3x and you end up with\[24x=144\]
then simplify and then you get\[x=6\]

Answer 31

thats right, isnt it :)

Answer 32

uhm, yea, not how I think about it but that works

Answer 33

also, if you divide by 3x the x is on the bottom, check your algebra

Answer 34

oh, right..in fixed it

Answer 35

show me again

Answer 36

\[2^3=144\div3x\] multiply both sides by the inverse of 3x\[(1/3x)*2^3=144\div3x*(1/3x)\]
im pretty sure this is correct
because then the x wont be on the bottom right?

Answer 37

no x is still on the bottom

Answer 38

argh im confused

Answer 39

how about get rid of the numbers and leave the x alone?

Answer 40

ie you have 4=6v solve for v what do you do?

Answer 41

okay...so um\[8=144\div3x\]\[8=(144\div3)x?\]

Answer 42

yup

Answer 43

now get rid of the blah in front of the x