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

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Question

(d)

Given, $72^x \cdot 48^y=6^{x y}$

$\left(2^3 \cdot 3^2\right)^x \cdot\left(2^4 \cdot 3\right)^y=2^{x y} \cdot 3^{x y}$

$2^{3 x+4 y} \cdot

Vedclass · Accepted Answer

$-\frac{10}{3}$

Vedclass · Answer

(d)

Given, $72^x \cdot 48^y=6^{x y}$

$\left(2^3 \cdot 3^2ight)^x \cdot\left(2^4 \cdot 3ight)^y=2^{x y} \cdot 3^{x y}$

$2^{3 x+4 y} \cdot 3^{2 x+y}=2^{x y} \cdot 3^{x y}$

Equating the exponent of $2$ and $3$ , we get

$3 x+4 y=x y$ and $2 x+y=x y$

On solving these equation, we get

$x=\frac{-15}{3}$ and $y=\frac{5}{3}$

$	herefore \quad x+y=\frac{-15}{3}+\frac{5}{3}=\frac{-10}{3}$

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

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