# ENIGMA Forums

## General fluff => Off-Topic => Topic started by: score_under on June 09, 2010, 11:17:28 AM

Title: logy(x) ≢ log(x)÷log(y)
Post by: score_under on June 09, 2010, 11:17:28 AM
logy(x) ≢ log(x) ÷ log(y)
Because:
log-1(-1) = Any odd integer, including negatives. (As (-1)-1 == -1, and (-1)1 == -1, etc.)
log(-1) ÷ log(-1) = 1.

WHAT.
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: Rusky on June 09, 2010, 11:53:27 AM
The domain of logs with non-negative bases is (0, ∞), so log(-1) doesn't exist (or is complex). You are getting part of the answer set, so I imagine either the log base conversion rule is only for positive bases or it's similar to trig problems where you have to deal with different periods.
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: score_under on June 09, 2010, 12:06:29 PM
so log(-1) doesn't exist (or is complex).
Either way, z/z ≡ 1 still holds in the complex plane (except for z=0, of course).
Quote
so I imagine either the log base conversion rule is only for positive bases
Perhaps so. I think teachers should specify this before teaching it to us.
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: Josh @ Dreamland on June 09, 2010, 12:28:17 PM
Eh, Wolfram reports log -1 of -1 is 1.
What told you it wasn't?

Oh, I see. This is the same problem you face with arcsin(), though. The method holds; there are just more answers than you are given by the function. It's not like this only happens with those, either; you see the same problem when you have x**2 = y. Because the sqrt() function only returns positive. As do all the other pow(x,1/n) functions, even though there are always n answers.
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: IsmAvatar on June 09, 2010, 12:57:24 PM
Thus: 1 = -1
We can further derive that, by adding 1 to each side and dividing by 2,
1 = 0
and a great number of math problems that I got the answers wrong on on tests can thusly be demonstrated as equivalent to the correct answer using this logic.
For any incorrect answer X, and the correct answer Y, it can be demonstrated that X = Y thusly:

1 = 0
1(X - Y) = 0(X - Y)
X - Y = 0

Or for those if you who like over-extraneous proofs

1: 1 = 0 (given)
2: 1X = 0X (multiply step 1 by X)
3: 1Y = 0Y (multiply step 1 by Y)
4: 0X = 0 (known - simplification)
5: 0Y = 0 (known - simplification)
6: 0 = 0 (known)
7: 0X = 0Y (derived by taking steps 4 and 5 and substituting in step 6)
8: 1X = 1Y (substitute steps 2 and 3 in step 7)
9: X = Y (simplify)
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: MahFreenAmeh on June 09, 2010, 06:01:38 PM
prove 1 without resorting to assigning a constant to a constant
Title: Re: logy(x) ≢ log(x)÷log(y)
Post by: Game_boy on June 10, 2010, 09:28:38 AM
Logs of negative numbers are complex (and have multiple answers as Rusky said). Added to the fact that complex number arithmetic doesn't follow the same rules, and you can 'derive' some strange results. In particular, all the 'log rules' don't necessarily apply and you have to re-derive them from first principles starting from 'what e^x means with complex arguments'.

But it is mathematically consistent.

And, yes, as Ism has done, you can prove all things starting from 1 = 0, but you don't get useful results.