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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

then













because



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)

 

#6 2006-09-06 01:15:28

Devantè
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Re: Logarithm Formulas










Last edited by Devanté (2006-10-07 21:23:57)

 

#7 2006-10-10 22:22:34

Devantè
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Re: Logarithm Formulas

http://www.mathsisfun.com/forum/latex/3/5/a1aa7d5f1f8b2b28b8366cf3688a7e1.gif

 

#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.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#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

 

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