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Name: Anonymous 2007-11-24 17:52

|log 2 (3x-1/3)|=|log 2 (5-2x/2)|
How does one go about solving such an equation?

Name: Anonymous 2007-11-24 18:10

log|2(3x-1/3)|=log|2(5-2x/2)|

10^log|2(3x-1/3)| = 10^log|2(5-2x/2)|
2(3x-1/3) = 2(5-2x/2)
3x-1/3 = 5-x
4x = 16/3
x= 4/3

Name: Anonymous 2007-11-24 18:29

>>2

protip, |log x| is not equal to log |x| in general.  like for any x in (0,1)

also, i think OP is noting log base 2 not log of 2 * some stuff. but eh.

Name: Anonymous 2007-11-24 18:32

>>2
Uhm, no, you're doing it wrong.
http://en.wikipedia.org/wiki/Absolute_value
And the 2 there is base 2 ... logarithm base 2 of (3x-1/3) and so on.

Name: Anonymous 2007-11-24 18:39

It's practically the same answer, no?

2^|log 2 (3x-1/3)| = 2^|log 2 (5-2x/2)|
3x-1/3 = 5-2x/2
etc.

Name: Anonymous 2007-11-24 18:53

>>5
If you forget about there being some ||||'s, yes.

Name: Anonymous 2007-11-24 19:00

2^|log 2 (3x-1/3)| = 2^|log 2 (5-2x/2)|
|3x-1/3| = |5-x|
|4x| = |16/3|
|x| = |4/3|

Name: Anonymous 2007-11-24 19:02

(3x - 1)/3 = (5-2x)/2

(3x - 1)/3 = 2/(5-2x)

3/(3x - 1) = (5-2x)/2

3/(3x - 1) = 2/(5-2x)

Solve each one, check the answers. The valid solutions are your answers.

Name: Anonymous 2007-11-24 19:07

Unless you actually meant 3x-1/3 and 5 - 2x/2
Which are (9x - 1)/3 and (5 - x)

Then solve

(9x - 1)/3 = (5 - x)
(9x - 1)/3 = 1/(5 - x)
1/(9x - 1) = 1/(5 - x)
1/(9x - 1) = (5 - x)

Check the solutions, etc

Name: Anonymous 2007-11-24 19:23

how come

(5 - x) =  1/(5 - x)
or
(5-2x)/2 =  2/(5-2x)
?

Name: Anonymous 2007-11-24 19:29

You have to check +-log 2 (3x-1/3) = +-log 2 (5-2x/2)|

Thus four possible solutions.  If either side is negative it goes to become the power of the inside and then you can set the insides equal.  and (5-x)^-1 is 1/(5-x)

Name: Anonymous 2007-11-24 19:37

| and (5-x)^-1 is 1/(5-x)

I know, but it says (5-x) not (5-x)^-1
it's like saying 42 = 1/42 ...

Anyway, I feel like a moron now... can't say I've had this math in school, but still...

Does anon care to explain every step of solving this equation to a fellow anon and maybe eligthen him a bit?

Name: Anonymous 2007-11-24 19:50

Since you've got +-log 2 (3x-1/3) = +-log 2 (5-2x/2) you need to solve

+log 2 (3x-1/3) = +log 2 (5-2x/2)
3x-1/3 = 5-2x/2

+log 2 (3x-1/3) = -log 2 (5-2x/2)
3x-1/3 = (5-2x/2)^-1

-log 2 (3x-1/3) = +log 2 (5-2x/2)
(3x-1/3)^-1 = 5-2x/2

-log 2 (3x-1/3) = -log 2 (5-2x/2)
(3x-1/3)^-1 = (5-2x/2)^-1

Name: Anonymous 2007-11-24 19:51

"it's like saying 42 = 1/42 ..."
Also, |log(42)| = |log(1/42)|

Name: Anonymous 2007-11-24 19:56

d'oh

(9x - 1)/3 = (5 - x)
(9x - 1)/3 = 1/(5 - x)
1/(9x - 1) = 1/(5 - x)
1/(9x - 1) = (5 - x)

is four different sets >.<

Name: Anonymous 2007-11-24 20:01

| |log(42)| = |log(1/42)|

Let's see if I understand this now...

+log(x) = -log(1/x)

-log(x) = +log(x^-1)

-log(x) =/= -log(x^-1)

+log(x) =/= +log(1/x)

so || means +- in a way...?

Name: Anonymous 2007-11-24 20:06

"so || means +- in a way...?"

If you were solving |x| = 2, then x could be either 2 or -2 because whatever is inside the absolute value signs could be either positive or negative, you don't know because the absolute value gets rid of any signs.  So to solve it you take +-x = 2, thus x=2 and x=-2, just like you get with basic intuition.


"Let's see if I understand this now...

+log(x) = -log(1/x)

-log(x) = +log(x^-1)

-log(x) =/= -log(x^-1)

+log(x) =/= +log(1/x)"

Yes

Name: Anonymous 2007-11-24 20:21

I've been enlightened, thanks anon.

Name: Anonymous 2007-11-25 3:17

>>3

Plus there's never any need to say |log(x)| since logs can never give negative values

Name: Anonymous 2007-11-25 13:33

>>19
Uh, log(0.5)

Name: Anonymous 2007-11-25 13:37

square both sides then solve

Name: Anonymous 2007-11-25 14:37

>>19

lol wut

you can never pass a negative number to a log, but the log of anything on (0,1) is negative.
Don't change these.
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