If {a^x} = {b^y} = {(ab)^{xy}},then x + y = | vedclass

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(B) Given that ${a^x} = {b^y} = {(ab)^{xy}}$.Let ${a^x} = {b^y} = {(ab)^{xy}} = k$.Then $a^x = k \implies a = k^{1/x}$ and $b^y = k \implies b = k^{1/

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$1$

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(B) Given that ${a^x} = {b^y} = {(ab)^{xy}}$.Let ${a^x} = {b^y} = {(ab)^{xy}} = k$.Then $a^x = k \implies a = k^{1/x}$ and $b^y = k \implies b = k^{1/y}$.Also,$(ab)^{xy} = k \implies ab = k^{1/xy}$.Substituting the values of $a$ and $b$ in the last equation:$k^{1/x} \cdot k^{1/y} = k^{1/xy}$$k^{(1/x + 1/y)} = k^{1/xy}$Equating the exponents:$\frac{1}{x} + \frac{1}{y} = \frac{1}{xy}$$\frac{x+y}{xy} = \frac{1}{xy}$Since $x, y 
eq 0$,we get $x + y = 1$.

If ${a^x} = {b^y} = {(ab)^{xy}}$,then $x + y = $

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