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#1 2008-09-15 13:04:07
- abotaweeela
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- Registered: 2008-09-15
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try it....!
X 2
=
2 X
X to power 2 = 2 to power X
X={2,4} try to proof that if you can?
Last edited by abotaweeela (2008-09-15 13:16:56)
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#2 2008-09-15 13:10:30
- Ricky
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Re: try it....!
Is that:
?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#3 2008-09-15 13:12:48
- abotaweeela
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Re: try it....!
no.... its not like that.
X to power 2 =2 to power X
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#4 2008-09-15 14:35:34
- Ricky
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Re: try it....!
Ah:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#5 2008-09-16 00:49:31
- mathsyperson
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Re: try it....!
There's another one: x ≈ -0.7667.
Those are the only three though.
Why did the vector cross the road?
It wanted to be normal.
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#6 2008-09-16 05:18:42
- TheDude
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- Registered: 2007-10-23
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Re: try it....!
I'm doing something wrong here, can anyone spot it?
Start by taking the log to remove the exponents. Let's use base 2 for simplicity's sake. For now we'll confine ourselves to values of x strictly greater than 0 (it's quickly proven that 0 is not a solution).
Now take the derivative of the left hand side to determine its critical points.
This would make x = 2 the only critical point, which can't be the case because both x = 2 and x = 4 are solutions.
Wrap it in bacon
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#7 2008-09-16 05:53:01
- Ricky
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Re: try it....!
Critical points don't correspond to 0's. In general, there is no (analytic) method of finding 0's. You just happened to get lucky that a critical point was also at a 0.
Edit after mikau's post: Really lucky.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#8 2008-09-16 05:53:11
- mikau
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- Registered: 2005-08-22
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Re: try it....!
i THINK what you're doing is saying, let y = x^2 - 2^x and trying to minimize y so that they become equal. but.. minimizing doesn't find the pair closest to zero. For that, you'd need to minimize the absolute value of y, and test to see if its zero.
also, looks like you forgot that the derivative of log2(x) is not 1/x, its 1/(ln(2)x ).
Last edited by mikau (2008-09-16 05:59:40)
A logarithm is just a misspelled algorithm.
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#9 2008-09-16 06:02:27
- TheDude
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- Registered: 2007-10-23
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Re: try it....!
Ahhh, mikau found it.
What I was implying originally is that x = 2 is a critical point, making it a local max or min, in this case a local max. If that was correct, then there would be no way for both x = 2 and x = 4 to be solutions to the equation, since f(4) would be strictly less than f(2). However, x = 2 is not a local max, x = 2^(1 / ln 2) is.
Last edited by TheDude (2008-09-16 06:05:23)
Wrap it in bacon
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