Q179
ArithmeticAlgebra, Equations & Operations
$216^{2/3} \div \left(\dfrac{P^2}{9}\right)^{-3/2} = 2^x \cdot 3^y \cdot P^z$. Find the values of $x$, $y$, $z$.
- a.x=2, y=-1, z=3
- b.x=-1, y=2, z=3
- c.x=3, y=-1, z=2
- d.x=2, y=3, z=-1
Answer: (A) x=2, y=-1, z=3
$216^{2/3} = 36$, and dividing by $\left(\dfrac{P^2}{9}\right)^{-3/2} = \dfrac{27}{P^3}$ gives $36 \times \dfrac{P^3}{27} = \dfrac{4}{3}P^3 = 2^2 \cdot 3^{-1} \cdot P^3$, so $x=2$, $y=-1$, $z=3$.