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Question Number 3239 by Rasheed Soomro last updated on 08/Dec/15

I {d o n}^{'} t k n o w t h e v a l u e o f L o g (- 1) b u t I c a l c u l a t e

i t i n t h e f o l l o w i n g w a y :

{(- 1)}^{2} = 1

\Rightarrow L o g {(- 1)}^{2} = L o g (1)

\Rightarrow 2 \times L o g (- 1) = 0

\Rightarrow L o g (- 1) = \frac{0}{2} = 0

A m I c o r r e c t ? I f n o, w h y ?

Commented by prakash jain last updated on 08/Dec/15

y = e^{x} \Rightarrow x = \ln y

e^{i x} = - 1 \Rightarrow \cos x + i \sin x = - 1

x = (2 n + 1) π

\ln (- 1) = i (2 n + 1) π

Commented by prakash jain last updated on 08/Dec/15

a^{2} = b^{2} ⇏ a = b

Commented by Rasheed Soomro last updated on 08/Dec/15

B u t w e h a v e n o t e x t r a c t e d s q u a r e r o o t o f b o t h s i d e s .

Commented by prakash jain last updated on 08/Dec/15

a = b

\ln {(\sqrt{a})}^{2} = \ln b

2 \ln \sqrt{a} = \ln b

\ln \sqrt{a} = \frac{1}{2} \ln b = \ln \sqrt{b}

a = b = 1

LHS = \sqrt{a} = - 1

RHS = \sqrt{b} = 1

Commented by Rasheed Soomro last updated on 08/Dec/15

O k S i r!

Commented by prakash jain last updated on 09/Dec/15

\ln 1 = 2 n π i

\ln (- 1) = (2 m + 1) i π

2 \ln (- 1) = 2 (2 m + 1) i π = 2 n π

This is correct result .

Commented by 123456 last updated on 08/Dec/15

e^{2 π ı} = e^{0} but 2 π ı \neq 0

Commented by 123456 last updated on 08/Dec/15

e^{x} = e^{y} \Leftrightarrow x = y work only for (x, y) \in R^{2}

at complex plane e^{z} dont is one to one

in fact

e^{z + 2 π n ı} = e^{z} n \in Z

Commented by Rasheed Soomro last updated on 08/Dec/15

T Han K^{S} S_{S!}!^{!}

Commented by 123456 last updated on 09/Dec/15

just random comment

\ln 1 = 2 π n ı

2 \ln (- 1) = 2 (π ı + m π ı) = 2 π (1 + m) ı

\ln {(- 1)}^{2} = 2 \ln (- 1)

n = 1 + m

\ln z = \ln ∣ z ∣ + ı [\arg (z) + 2 π k]

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