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Question Number 68712 by Rio Michael last updated on 15/Sep/19

g i v e n t h a t x a n d y a r e t w o n u m b e r s o t h e r o n e .

g i v e n t h a t a > 0 a n d b > 0

a n d a^{x} = b^{y} = {(a b)}^{x y} s h o w t h a t x + y = 0

Commented by Prithwish sen last updated on 15/Sep/19

Let a^{x} = b^{y} = {(ab)}^{xy} = k

∴ a = k^{\frac{1}{x}}

b = k^{\frac{1}{y}}

ab = k^{\frac{1}{xy}}

∴ k^{\frac{1}{x}} . k^{\frac{1}{y}} = k^{\frac{1}{xy}} \Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{xy} \Rightarrow x + y = 1

please check

Commented by MJS last updated on 15/Sep/19

a^{0} + b^{0} = {(a b)}^{0} \Rightarrow x = y = 0

is one solution

a^{x} = b^{y} \Rightarrow y = \frac{\ln a}{\ln b} x

a^{x} = {(a b)}^{x y} \Rightarrow y = \frac{\ln a}{\ln a + \ln b}

\frac{\ln a}{\ln b} x = \frac{\ln a}{\ln a + \ln b} \Rightarrow x = \frac{\ln b}{\ln a + \ln b}

\Rightarrow x + y = 1

Commented by Prithwish sen last updated on 15/Sep/19

Thanks Sir .