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

3

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\frac{10}{3}

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3

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- \frac{10}{3}

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Solution

The correct option is D $- \frac{10}{3}$
$72^{x} {.48}^{y} = 6^{x y} \Rightarrow (2^{3} \times 3^{2})^{x} . (2^{4} \times 3^{1})^{y} = 2^{x y} \times 3^{x y} \Rightarrow 2^{3 x + 4 y} \times 3^{2 x + y} = 2^{x y} \times 3^{x y}$
So, $3 x + 4 y = x y \dots (1)$
$2 x + y = x y \dots (2)$

From $(1)$ and $(2),$ we get
$x = - 3 y$
Now, $- 9 y + 4 y = - 3 y^{2}$
$\Rightarrow 3 y^{2} - 5 y = 0 \Rightarrow y = \frac{5}{3} [∵ y is non zero]$
$∴ x = - 5$
Hence, $x + y = - \frac{10}{3}$

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If 72x.48y=6xy, where x and y are non zero rational numbers, then x+y equals

The correct option is D −103 72x.48y=6xy⇒(23×32)x.(24×31)y=2xy×3xy⇒23x+4y×32x+y=2xy×3xy So, 3x+4y=xy ⋯(1) 2x+y=xy ⋯(2) From (1) and (2), we get x=−3y Now, −9y+4y=−3y2 ⇒3y2−5y=0⇒y=53 [∵y is non zero ] ∴x=−5 Hence, x+y=−103

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