which expression is equivalent to $sqrt4{\frac{16x^{11}y^{8}}{81x^{7}y^{6}}}$? assume $x > 0$ and…

which expression is equivalent to $sqrt4{\frac{16x^{11}y^{8}}{81x^{7}y^{6}}}$? assume $x > 0$ and $y\neq0$.\n$\frac{4x(sqrt4{y^{2}})}{9}$\n$\frac{2x(sqrt4{y^{2}})}{3}$\n$\frac{4x^{2}y}{9}$\n$\frac{2x^{2}y}{3}$

which expression is equivalent to $sqrt4{\frac{16x^{11}y^{8}}{81x^{7}y^{6}}}$? assume $x > 0$ and $y\neq0$.\n$\frac{4x(sqrt4{y^{2}})}{9}$\n$\frac{2x(sqrt4{y^{2}})}{3}$\n$\frac{4x^{2}y}{9}$\n$\frac{2x^{2}y}{3}$

Answer

Explanation:

Step1: Simplify the fraction inside the fourth - root

Use the quotient rule of exponents $\frac{a^m}{a^n}=a^{m - n}$ and $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$. $\sqrt[4]{\frac{16x^{11}y^{8}}{81x^{7}y^{6}}}=\frac{\sqrt[4]{16x^{11}y^{8}}}{\sqrt[4]{81x^{7}y^{6}}}$ Since $\sqrt[4]{16} = 2$ and $\sqrt[4]{81}=3$, and using the product rule of exponents $\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}$, we have $\frac{\sqrt[4]{16}\cdot\sqrt[4]{x^{11}}\cdot\sqrt[4]{y^{8}}}{\sqrt[4]{81}\cdot\sqrt[4]{x^{7}}\cdot\sqrt[4]{y^{6}}}=\frac{2\sqrt[4]{x^{11}}\cdot y^{2}}{3\sqrt[4]{x^{7}}\cdot\sqrt[4]{y^{6}}}$

Step2: Simplify the $x$ - terms using the quotient rule of exponents

$\frac{\sqrt[4]{x^{11}}}{\sqrt[4]{x^{7}}}=x^{\frac{11}{4}-\frac{7}{4}}=x^{\frac{11 - 7}{4}}=x^{1}$ The $y$ - terms: $\frac{y^{2}}{\sqrt[4]{y^{6}}}=y^{2-\frac{6}{4}}=y^{2 - \frac{3}{2}}=y^{\frac{4 - 3}{2}}=y^{\frac{1}{2}}=\sqrt{y}$ (but we can also simplify the original fourth - root in another way) Another way: $\sqrt[4]{\frac{16x^{11}y^{8}}{81x^{7}y^{6}}}=\sqrt[4]{\frac{16}{81}\cdot x^{11 - 7}\cdot y^{8 - 6}}=\sqrt[4]{\frac{16}{81}\cdot x^{4}\cdot y^{2}}$ $=\sqrt[4]{\frac{16}{81}}\cdot\sqrt[4]{x^{4}}\cdot\sqrt[4]{y^{2}}=\frac{2x\sqrt[4]{y^{2}}}{3}$

Answer:

$\frac{2x\sqrt[4]{y^{2}}}{3}$