- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
normal
If $72^x \cdot 48^y=6^{x y}$, where $x$ and $y$ are non-zero rational numbers, then $x+y$ equals
A
$3$
B
$\frac{10}{3}$
C
$-3$
D
$-\frac{10}{3}$
(KVPY-2017)
Solution
(d)
Given, $72^x \cdot 48^y=6^{x y}$
$\left(2^3 \cdot 3^2\right)^x \cdot\left(2^4 \cdot 3\right)^y=2^{x y} \cdot 3^{x y}$
$2^{3 x+4 y} \cdot 3^{2 x+y}=2^{x y} \cdot 3^{x y}$
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 $y=\frac{5}{3}$
$\therefore \quad x+y=\frac{-15}{3}+\frac{5}{3}=\frac{-10}{3}$
Standard 11
Mathematics
