If 72^x cdot 48^y = 6^{xy},where x and y are | vedclass

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(D) Given,$72^x \cdot 48^y = 6^{xy}$.Expressing the bases as powers of $2$ and $3$:$(2^3 \cdot 3^2)^x \cdot (2^4 \cdot 3^1)^y = 2^{xy} \cdot 3^{xy}$.$

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$-\frac{10}{3}$

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(D) Given,$72^x \cdot 48^y = 6^{xy}$.Expressing the bases as powers of $2$ and $3$:$(2^3 \cdot 3^2)^x \cdot (2^4 \cdot 3^1)^y = 2^{xy} \cdot 3^{xy}$.$2^{3x+4y} \cdot 3^{2x+y} = 2^{xy} \cdot 3^{xy}$.Equating the exponents of $2$ and $3$ on both sides:$3x + 4y = xy$  $(1)$$2x + y = xy$  $(2)$From $(1)$ and $(2)$,we have $3x + 4y = 2x + y$,which simplifies to $x = -3y$.Substitute $x = -3y$ into equation $(2)$:$2(-3y) + y = (-3y)y$$-6y + y = -3y^2$$-5y = -3y^2$.Since $y 
eq 0$,we can divide by $y$:$-5 = -3y \implies y = \frac{5}{3}$.Now,find $x$:$x = -3 \left(\frac{5}{3}ight) = -5$.Therefore,$x + y = -5 + \frac{5}{3} = \frac{-15 + 5}{3} = -\frac{10}{3}$.

If $72^x \cdot 48^y = 6^{xy}$,where $x$ and $y$ are non-zero rational numbers,then $x+y$ equals

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