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

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(B) Given: ${a^x} = {b^y} = {(ab)^{xy}}$Taking natural logarithm on all sides:$x \ln a = y \ln b = xy \ln (ab) = k$ (where $k$ is a constant)From this

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(B) Given: ${a^x} = {b^y} = {(ab)^{xy}}$Taking natural logarithm on all sides:$x \ln a = y \ln b = xy \ln (ab) = k$ (where $k$ is a constant)From this,we get:$\ln a = \frac{k}{x}$ and $\ln b = \frac{k}{y}$Also,$\ln (ab) = \frac{k}{xy}$Since $\ln (ab) = \ln a + \ln b$,we substitute the values:$\frac{k}{x} + \frac{k}{y} = \frac{k}{xy}$Dividing both sides by $k$ (assuming $k 
eq 0$):$\frac{1}{x} + \frac{1}{y} = \frac{1}{xy}$$\frac{x + y}{xy} = \frac{1}{xy}$Therefore,$x + y = 1$.

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

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