# ENIGMA Development Environment

 Pages: 1
 Author Topic: logy(x) ≢ log(x)÷log(y)  (Read 1390 times)
score_under
 Posted on: June 09, 2010, 11:17:28 AM

Joined: Aug 2008
Posts: 308

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.
 « Last Edit: June 09, 2010, 11:31:00 AM by score_under » Logged
Rusky
 Reply #1 Posted on: June 09, 2010, 11:53:27 AM

Joined: Feb 2008
Posts: 955

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.
 Logged
score_under
 Reply #2 Posted on: June 09, 2010, 12:06:29 PM

Joined: Aug 2008
Posts: 308

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.
 Logged
Josh @ Dreamland
 Reply #3 Posted on: June 09, 2010, 12:28:17 PM

Prince of all Goldfish

Location: Pittsburgh, PA, USA
Joined: Feb 2008
Posts: 2959

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.
 « Last Edit: June 09, 2010, 12:31:15 PM by Josh @ Dreamland » Logged
"That is the single most cryptic piece of code I have ever seen." -Master PobbleWobble
"I disapprove of what you say, but I will defend to the death your right to say it." -Evelyn Beatrice Hall, Friends of Voltaire
IsmAvatar
 Reply #4 Posted on: June 09, 2010, 12:57:24 PM

LateralGM Developer

Location: Pennsylvania/USA
Joined: Apr 2008
Posts: 886

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)
 Logged
MahFreenAmeh
 Reply #5 Posted on: June 09, 2010, 06:01:38 PM

"Web Team"
Location: Austin, TX
Joined: Apr 2008
Posts: 13

prove 1 without resorting to assigning a constant to a constant
 Logged
[java,c++,c,javascript,html,css,php,perl,ruby,python,sql]
if you've got ideas, let me hear them.
Game_boy
 Reply #6 Posted on: June 10, 2010, 09:28:38 AM

Joined: Apr 2008
Posts: 228

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.
 Logged
 Pages: 1