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Can this be solved fully algebraically?
2^x = 2x
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Wierd. The answer is obviously x = 2 but I can't seem to solve it using logarithms or anything like that.
A logarithm is just a misspelled algorithm.
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One thing I noted is if x^n = n, then x is then x is the nth root of n. (you can prove this by raising both sides to the 1/nth power. But I don't think that will help here.
A logarithm is just a misspelled algorithm.
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Hmm. I don't really know what I'm doing here, but I'll give it a go.
2^x = 2x
2^(x-1) = x
x-1 = lg(2) x
...and I don't know after that. Bleh.
x = 1 would also be an answer.
Why did the vector cross the road?
It wanted to be normal.
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What about the inverses of the functions?
If you combine mathsypersons stuff with mine,
you get x-1 = x/2, but I don't know if that's legal
after I took the inverse.
Last edited by John E. Franklin (2005-12-21 11:06:27)
igloo myrtilles fourmis
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Yeah I tried logs and changes of bases and etc but it just simplifies to the original equation.
The only answers are x = 1 and x = 2, which can be proved once you do "find" them...
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Let lg2 represent logarithm to the base 2.
2^x=2x
Taking lg2 on both sides,
x = lg2(2x) = lg2(2) + lg2(x)
lg2(2)=1,
Therefore,
x = 1+ lg2(x)
x-1 = lg2(x)
Raising both sides to the power 2,
2^(x-1)=x
For 1 and 2 alone, the LHS=RHS.
Mathsyperson was on the right path (as he's always )
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Hi.
I tried a thing like this (but I think that it's the same thing)
Let ln log base e
2^x=2x
ln(2^x)=ln2+lnx
xln2=ln2+lnx
x=1+ln(x)/ln(2)
and so
x=1+Log (x)
2
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brilliant work, guys!
Logarithms rock!
A logarithm is just a misspelled algorithm.
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at the end of post#5 I had x-1 = x/2, but is this allowed since
I took the inverse of the functions?
igloo myrtilles fourmis
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It is definately true that it only works for 1 and 2, but I guess, being more specific, is it possible to isolate x? As in use algebra to bring the equation down to an x = 1?
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We have two answers:
x=1 and x=2.
IPBLE: Increasing Performance By Lowering Expectations.
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Geometric proof:
Last edited by krassi_holmz (2005-12-28 09:53:50)
IPBLE: Increasing Performance By Lowering Expectations.
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And we'll proof that there doesn't exist line that intersects with 2^i more than 2 times. That's because
ln x means natural logaritm of x
(2^i)'=2^i(ln2)=2^iC
That means that (2^i)' is a monotonic growing function so (2^i) is "convex".
IPBLE: Increasing Performance By Lowering Expectations.
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Condition for existing line that divides grafhic of f(x) more than 2 times is that is nessesery to exist at least one inflex point.
IPBLE: Increasing Performance By Lowering Expectations.
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Could somebody proof the upper using analisis?
I'll try solving it but i'm not so good at analisis.
IPBLE: Increasing Performance By Lowering Expectations.
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