# Am i a genious or a complete idiot?Watch

13 years ago
#41
and yet you were conviced that root2root2root2 = 3root2....

none the less, congrats on your offer.
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13 years ago
#42
I saw a far better 'fake proof' today - which showed that all triangles are actually isoscles.
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#43
oki, but let me tell u something more amusing now.

my maths teacher used to tell me that there couldn't be a log of a negative number. but look at this:

If a^b = c and then log(base a)c = b

say 1^-5 = 1 then log(base1) -5 = 1

so now tell me WHATS WRONG WITH THAT????? IS MY MATHS TEACHER STUPID OR WHA???
-----> (maybe he's a geniOUS too ..lol )
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13 years ago
#44
the base of log must be > 1

--------------

edit <> 1, not > 1
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13 years ago
#45
[QUOTE=Candy-Kills]
say 1^-5 = 1 then log(base1) -5 = 1
QUOTE]

Dont you mean log(base1) 1 = -5?
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#46
oh, ****! i meessed me up!! :O

--------------

ill try again later, give me some time to remember how it was...
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13 years ago
#47
Don't try to dumb yourself, as logarithm is not defined for base = 1
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13 years ago
#48
that's interesting i didn't know that...
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13 years ago
#49
A far better example of what you're trying to show would be:

x2 = (-x) 2
=> log(x2)=log((-x) 2 )
=> 2logx=2log (-x)
=> logx=log(-x)

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13 years ago
#50
(Original post by Gemini)
A far better example of what you're trying to show would be:

x2 = (-x) 2
=> log(x2)=log((-x) 2 )
=> 2logx=2log (-x)
=> logx=log(-x)

I don't see anything wrong there.

(Unless you're saying that implies x=-x. )
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13 years ago
#51
Lol... rename this thread ppl!.... geek heaven. I feel right at home
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13 years ago
#52
(Original post by Gemini)
A far better example of what you're trying to show would be:

x2 = (-x) 2
=> log(x2)=log((-x) 2 )
=> 2logx=2log (-x)
=> logx=log(-x)

Infact you are using log(ab) = loga + logb
The domain of logarithm is also positive.
=> It's only be defined with a, b > 0
If you have ab > 0 but a < 0, b < 0
then log(ab) is still defined, but loga and logb are not
And you must write
log(ab) = log|a| + log|b|
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13 years ago
#53
Taking the branch cut to be the principle one (ie down the negative real axis) the standard definition of the logarithm of a negative number is :

Gah, it's as if noone has heard of complex methods or residue calculus
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13 years ago
#54
oooh my goodness ok u people are scaring me now
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13 years ago
#55
(Original post by AlphaNumeric)
Taking the branch cut to be the principle one (ie down the negative real axis) the standard definition of the logarithm of a negative number is :

Gah, it's as if noone has heard of complex methods or residue calculus
first time I see it, but anyways, it can't be iπ = 0, right? Hehe ... another idiot here if I say iπ = 0?
Is it because of log(x) and log(-x) are in different sets?
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13 years ago
#56
Have a gander at http://mathworld.wolfram.com/NaturalLogarithm.html specifically Equation 8.

That is true for all complex numbers except those on the negative real axis.
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13 years ago
#57
(Original post by AlphaNumeric)

That is true for all complex numbers except those on the negative real axis.
Didn't your post a few lines up just show that it was true for negative reals? Specifically, arg(-x) = pi, and ln|-x| = ln x

Edit: Oh, I see that you took your branch cut to be along the -ve real axis. That explains why the function's not defined there. Intellectual honesty compels me to point out that ln(z) is multivalued, a fact which you glossed over with the simple phrase "principal branch"
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#58
ermmmm... yeah..sure maybe when i finish my degree ill understand somethin of that
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13 years ago
#59
but if we use a multivariate ∫∑«σ≡Ø» analysis method we see that the moon is made of cheese.
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13 years ago
#60
(Original post by Candy-Kills)
oki, but let me tell u something more amusing now.

my maths teacher used to tell me that there couldn't be a log of a negative number. but look at this:

If a^b = c and then log(base a)c = b

say 1^-5 = 1 then log(base1) -5 = 1

so now tell me WHATS WRONG WITH THAT????? IS MY MATHS TEACHER STUPID OR WHA???
-----> (maybe he's a geniOUS too ..lol )
Now look at what you wrote a line before. It's log(base 1) 1 = -5, which is true though not really helpful.

Edit: OK, so in the time it took me to read this thread another page appeared. Oh well.
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