The-result-log-xy-log-x-log-y-is-not-always-true-Give-a-pair-of-values-of-x-and-y-such-that-the-result-will-not-hold-

Question Number 5569 by Rasheed Soomro last updated on 20/May/16

The result “ \log xy = \log x + \log y^{″} is not

always true! Give a pair of values of

x and y such that the result will not hold .

Commented by Yozzii last updated on 20/May/16

(x, y) = (- 1, 2)

\Rightarrow c o m p l e x r e s u l t o f l o g x y a n d l o g x w i t h

i n f i n i t e l y m a n y a n s w e r s .

S i n c e c o s n π = {(- 1)}^{n} n \in Z

\Rightarrow c o s (2 n - 1) π = {(- 1)}^{2 n - 1} = {(- 1)}^{2 n} \times (- 1) = - 1

∴ F o r a > 0, l n (- a) = l n ({a e}^{(2 n - 1) π i})

l n (- a) = l n a + (2 n - 1) π i n \in Z

l o g x y = l o g (- 2) = \frac{1}{l n 10} (l n 2 + (2 n - 1) π i) n \in Z

l o g (- 1) = \frac{1}{l n 10} (l n 1 + (2 j - 1) π i) j \in Z

l o g x = l o g (- 1) = \frac{(2 j - 1) π i}{l n 10}

l o g y = \frac{l n 2}{l n 10}

∴ l o g x + l o g y = \frac{1}{l n 10} (l n 2 + (2 j - 1) π i)

L e t j = 2 \Rightarrow l o g x + l o g y = \frac{1}{l n 10} (l n 2 + 3 π i)

L e t n = 1 \Rightarrow l o g x y = \frac{1}{l n 10} (l n 2 + π i)

∴ l o g x y = l o g x + l o g y

\Rightarrow \frac{1}{l n 10} (l n 2 + π i) = \frac{1}{l n 10} (l n 2 + 3 π i)

\Rightarrow 1 = 3 w h i c h i s i m p o s s i b l e .

Commented by Rasheed Soomro last updated on 20/May/16

TH α NKS!