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Logarithm Formulas
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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If
thenIt 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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Last edited by Ricky (2006-04-04 04:11:44)
"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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Logarithm of a Complex Number
where k is an integer.
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(which is log(xy) = log x + log y)
Last edited by Devanté (2006-08-05 19:10:52)
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Last edited by Devanté (2006-10-06 23:23:57)
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Can anyone explain Devante's post #7? He says
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Can anyone prove the following....
Given,
((2/3)^k)n = 1
Required to prove,
k is equal to log of n on base 3/2
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Hi random_fruit;
It could be a quiz problem and I agree it appears to have only 1 solution in R.
Last edited by bobbym (2009-04-19 19:44:38)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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If
then
because
hey ganesh jus a little thing
if we take A and B of the same signs
and in the other ones also... the argument has to be alwys positive
Everything that has a begining has an EnD!!!
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Hi Shekhar,
Here is the solution.
((2/3)^k)n = 1
⇒(2/3)^k = 1/n
⇒(2/3)^k = n-¹
Applying log on both sides.
⇒k log(2/3) = -log n
Multiplying (-) on both sides
⇒k log(2/3)-¹ = log n
⇒k log(3/2) = log n
∴ k = (log n)/(log (3/2))
⇒ k = log n base (3/2)
Can anyone prove the following....
Given,
((2/3)^k)n = 1
Required to prove,
k is equal to log of n on base 3/2
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