Chapter 7.4, Problem 71PPE

To determine

To simplify: the expression ${\left(\frac{3x{y}^5}{x^{4}y}\right)}^{-2}$

Answer to Problem 71PPE

The required answer is: $\frac{0.111x^6}{y^8}$

Explanation of Solution

Given information:

The given expression: ${\left(\frac{3x{y}^5}{x^{4}y}\right)}^{-2}$

Calculation:

  ${\left(\frac{3x{y}^5}{x^{4}y}\right)}^{-2}=\frac{1}{{\left(\frac{3x{y}^5}{x^{4}y}\right)}^2}$

(Rewrite using positive exponents)

  $=\frac{1}{\frac{{(3x{y}^5)}^2}{{(x^{4}y)}^2}}$

(Raise the quotient to the second power)

  $=\frac{1}{\frac{3^2{x}^2{y}^{10}}{x^8{y}^2}}$

(Multiply the exponents)

  $=1\cdot \frac{x^8{y}^2}{3^2{x}^2{y}^{10}}$

(Multiply by the reciprocal of $\frac{3^2{x}^2{y}^{10}}{x^8{y}^2}$ which is $\frac{x^8{y}^2}{3^2{x}^2{y}^{10}}$ )

  $=\frac{1}{3^2}{x}^{8-2}{y}^{2-10}$

(Subtract exponents when dividing powers with the same base)

  $\begin{array}{l}=\frac{1}{3^2}{x}^6{y}^{-8}\\ =\frac{1}{3^2}{x}^6\cdot \frac{1}{y^8}\end{array}$

(Rewrite using positive exponents)

  $\begin{array}{l}=\frac{1}{9}\cdot \frac{x^6}{y^8}\\ =\frac{0.111x^6}{y^8}\end{array}$

(Simplify)