Chapter 7, Problem 35CR

To determine

To fill: The blank and complete the equation ${\left(2x^2{y}^{\frac{1}{4}}\right)}^{?}=4x^4{y}^{\frac{1}{2}}$ .

Answer to Problem 35CR

The equation is ${\left(2x^2{y}^{\frac{1}{4}}\right)}^2=4x^4{y}^{\frac{1}{2}}$ .

Explanation of Solution

Given information:

Theequation is ${\left(2x^2{y}^{\frac{1}{4}}\right)}^{?}=4x^4{y}^{\frac{1}{2}}$ .

Calculation:

Consider the equation as ${\left(2x^2{y}^{\frac{1}{4}}\right)}^m=4x^4{y}^{\frac{1}{2}}$ .

Simplify the equation ${\left(2x^2{y}^{\frac{1}{4}}\right)}^m=4x^4{y}^{\frac{1}{2}}$ to obtain the value of mas follows:

  $\begin{array}{c}{\left(2x^2{y}^{\frac{1}{4}}\right)}^m=4x^4{y}^{\frac{1}{2}}\\ {\left(2x^2{y}^{\frac{1}{4}}\right)}^m=2\times 2\times x^{2\times 2}{y}^{\frac{1}{4}\times 2}\\ {\left(2x^2{y}^{\frac{1}{4}}\right)}^m=2^2{x}^{2\times 2}{y}^{\frac{1}{4}\times 2}\\ {\left(2x^2{y}^{\frac{1}{4}}\right)}^m={\left(2x^2{y}^{\frac{1}{4}}\right)}^2\end{array}$

Compare the exponentson both sides of the equation.

  $m=2$

Therefore, the value of mis 2.

Thus, the equation is ${\left(2x^2{y}^{\frac{1}{4}}\right)}^2=4x^4{y}^{\frac{1}{2}}$ .