logs

[Fearless™]

In my textbook, log law #3 states:
Logx^m = mlogx (x>0)
But take this for example:

Log [(x+2/x+4)]²
= 2log(x+2) – 2log(x+4)
Where x > -2


if you take –3, and substitute it in Log [(x+2/x+4)]², it works out. If you substitute it in 2log(x+2) – 2log(x+4), then it gives an error.

So my question is…is there something wrong with the 3rd law? Or what?

[Fearless™]

Originally Posted by LolSponge

Na you're just retarded.

answer the question or gtfo
*&&* I asked my dad...even he's clueless...

[Essayist]

Letting x = -3 will result in an error in one expression but not the other because the log function log(x) is defined only for values where x > 0.

log(x^m) = mlogx; x > 0

This reads that the log of x^m = mlogx only when x > 0.

If your expression inside the log function is (x+2)/(x+4) then the third law applies only when x + 2 > 0. If you let x = -3, you will find that -3 + 2 is not > 0, and so you cannot apply this law.

Remember that the law states x > 0; not x^m > 0. The third law of logarithms was not violated. Perhaps you just read it wrong. Deriving mlogx from log(x^m) is only possible if x > 0. If "x" in your case is (x+2)/(x+4) and your "m" is 2, then you can derive 2 log [(x+2)/(x+4)] only when (x+2)/(x+4) > 0. It doesn't matter if [(x+2)/(x+4)]^2 > 0 (in fact, [(x+2)/(x+4)]^2 is ALWAYS > 0 if x does not equal -2 or -4.)

In a nutshell, you misapplied the third law which resulted in a contradiction. In one expression, you found that log[(x+2)/(x+4)]^2 where x = -3 gave you a real answer. This is obvious because [(-3+2)/[-3+4)]^2 = 1 and log(1) is defined. Where you went wrong was assuming because log [(x+2)/(x+4)]^2 is defined the third law could be applied to derive 2 log [(x+2)/(x+4)]. See the problem with that? Now letting x = -3 will give you an unreal answer because log [(x+2)/(x+4)] is not defined at x = -3. This is because (x+2)/(x+4) MUST BE > 0 or else the third law is not applicable.

Remember that the law states log(x^m) = mlogx only when x > 0; not x^m > 0. The third law of logarithms was not violated.

[Fearless™]

Originally Posted by Essayist

Letting x = -3 will result in an error in one expression but not the other because the log function log(x) is defined only for values where x > 0.

log(x^m) = mlogx; x > 0

This reads that the log of x^m = mlogx only when x > 0.

If your expression inside the log function is (x+2)/(x+4) then the third law applies only when x + 2 > 0. If you let x = -3, you will find that -3 + 2 is not > 0, and so you cannot apply this law.

Remember that the law states x > 0; not x^m > 0. The third law of logarithms was not violated. Perhaps you just read it wrong. Deriving mlogx from log(x^m) is only possible if x > 0. If "x" in your case is (x+2)/(x+4) and your "m" is 2, then you can derive 2 log [(x+2)/(x+4)] only when (x+2)/(x+4) > 0. It doesn't matter if [(x+2)/(x+4)]^2 > 0 (in fact, [(x+2)/(x+4)]^2 is ALWAYS > 0 if x does not equal -2 or -4.)

In a nutshell, you misapplied the third law which resulted in a contradiction. In one expression, you found that log[(x+2)/(x+4)]^2 where x = -3 gave you a real answer. This is obvious because [(-3+2)/[-3+4)]^2 = 1 and log(1) is defined. Where you went wrong was assuming because log [(x+2)/(x+4)]^2 is defined the third law could be applied to derive 2 log [(x+2)/(x+4)]. See the problem with that? Now letting x = -3 will give you an unreal answer because log [(x+2)/(x+4)] is not defined at x = -3. This is because (x+2)/(x+4) MUST BE > 0 or else the third law is not applicable.

Remember that the law states log(x^m) = mlogx only when x > 0; not x^m > 0. The third law of logarithms was not violated.

Erm...some parts I understood...some parts confuzzled me...>.<

[Fearless™]

So, umm, does that mean I can't write Log [(x+2/x+4)]² as 2log(x+2) – 2log(x+4)?

Anyhow, here's the question: expand the following...
Log [(x+2/x+4)]²

So what's the answer?

edit: I promise to rep you later on =) I understand it now...

edit 2: are you prettz? :S

[Essayist]

To expand log [(x+2/x+4)]² you can still apply the third law, but only for x values where x > -2.

[Fearless™]

Originally Posted by Essayist

To expand log [(x+2/x+4)]² you can still apply the third law, but only for x values where x > -2.

Ohhhh, now I see. So if I were given [(x+2/x+4)]², (and not told to expand) then it is valid for x = R?

[The Voice]

Okay this thread just gave me a headache.

[Essayist]

Originally Posted by Fearless™

Ohhhh, now I see. So if I were given [(x+2/x+4)]², (and not told to expand) then it is valid for x = R?

You're almost right. log[(x+2)/(x+4)]² is defined for all x values except where x = -4 and x = -2.

[Fearless™]

Originally Posted by Essayist

You're almost right. log[(x+2)/(x+4)]² is defined for all x values except where x = -4 and x = -2.

omg omg. I KNEW that!!! Lololol. x___x I just forgot to put it...It's a bad habit really, I always forget to check the x-values @__@'; but thankies Prettz <3