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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,558

Logarithm Formulas

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,414

If

thenbecause

Character is who you are when no one is looking.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

*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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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

**Logarithm of a Complex Number**

where k is an integer.

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

(which is log(xy) = log *x* + log *y*)

*Last edited by Devanté (2006-08-05 19:10:52)*

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**Devantè****Real Member**- Registered: 2006-07-14
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*Last edited by Devanté (2006-10-06 23:23:57)*

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**Devantè****Real Member**- Registered: 2006-07-14
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**random_fruit****Member**- Registered: 2008-12-25
- Posts: 39

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.

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**Shekhar****Member**- Registered: 2009-04-19
- Posts: 1

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,680

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**noobard****Member**- Registered: 2009-06-07
- Posts: 28

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!!!

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**sekhar5955****Member**- Registered: 2009-11-25
- Posts: 1

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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