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#1 2006-03-30 09:04:14
- MathsIsFun
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Logarithm Formulas
Logarithm Formulas
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#2 2006-04-02 15:52:46
- ganesh
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Re: Logarithm Formulas
If
thenbecause
Character is who you are when no one is looking.
#3 2006-04-05 02:11:31
- Ricky
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Re: Logarithm Formulas
Last edited by Ricky (2006-04-05 02:11:44)
"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..."
#4 2006-08-06 11:53:15
- Zhylliolom
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Re: Logarithm Formulas
Logarithm of a Complex Number
where k is an integer.
#5 2006-08-06 17:03:45
- Devantè
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Re: Logarithm Formulas
(which is log(xy) = log x + log y)
Last edited by Devanté (2006-08-06 17:10:52)
#8 2009-01-03 05:03:12
- random_fruit
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Re: Logarithm Formulas
Can anyone explain Devante's post #7? He says
This does not make sense to me, and seems to me to have exactly one value of x for which it is true.
#9 2009-04-20 16:30:04
- Shekhar
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Re: Logarithm Formulas
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
#10 2009-04-20 17:39:47
- bobbym
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Re: Logarithm Formulas
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 (2009-04-20 17:44:38)
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#11 2009-06-10 19:04:36
- noobard
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Re: Logarithm Formulas
ganesh wrote:
If
then
because
hey ganesh jus a little thing
if we take A and B of the same signs
here it should be
..
and in the other ones also... the argument has to be alwys positive
Everything that has a begining has an EnD!!!
#12 2009-11-26 22:06:46
- sekhar5955
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Re: Logarithm Formulas
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)
Shekhar wrote:
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