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Algebra

Cool goIITian

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3 Aug 2009 22:32:10 IST

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CHALLANGE!!!!

None

Calculate log of (-e) to the base e????

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dc16

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3 Aug 2009 22:34:28 IST

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the domain of Log is from 0 to infinity

KALINDI

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3 Aug 2009 22:37:32 IST

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ITS DOMAIN IS >0..............SO INSIDE LOG SHLD BE >0

Shreyas Bhatt

Cool goIITian

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3 Aug 2009 22:43:03 IST

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come guys that's what is a challenge

Shreyas Bhatt

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3 Aug 2009 22:44:28 IST

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if u want I can give u hint.

KALINDI

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3 Aug 2009 22:47:11 IST

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OK GIVE THE HINT.........QUES HI GALAT HAI TOH HINT KAUNSA SAHI HOGA........

Shreyas Bhatt

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3 Aug 2009 23:16:21 IST

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The answer is imaginary

u r right that the domain of log is greater that zero which means that the answer is not real.

Let me explain u

domain of square root is also greater that zero but still we wite root of (-1) 'i' as an imaginary no.

KALINDI

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3 Aug 2009 23:30:19 IST

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SAB TUMHARE DIMAAG KI UPAJ HAI...........

Shreyas Bhatt

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3 Aug 2009 23:34:57 IST

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C'mon, the may be the first of its kind but it is as true as the complex no. are.

Shreyas Bhatt

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3 Aug 2009 23:35:50 IST

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I can even provide u the answer.

Bipin Dubey

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4 Aug 2009 05:49:49 IST

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It is not real but can be calculated in terms of complex numbers.

let log_e(-e) = x + iy ( x and y are real numbers)

e^x+iy = -e

e^x(cosy + isiny) = -e

equating real and imaginary parts : e^xcosy = -e and e^xsiny = 0

siny=0 ; cosy = -1 since e^x is +ve. So y = (2n+1)pi for any integre n

e^x = e => x=1

so, log_e(-e) = x + iy = 1 + i {(2n+1)pi}

Shreyas Bhatt

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4 Aug 2009 10:33:23 IST

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ya it is write and similarly we can go for log of any complex no.

Anant Kumar

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5 Aug 2009 08:06:11 IST

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$\ln(-e)=\ln(e^{i(2n+1)\pi}e})=\ln(e^{1+i(2n+1)\pi})=1+i(2n+1)\pi$

Here n is an integer.

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