Chapter 7.2, Problem 17CYU

To determine

To Simplify: The Given expression

  $\frac{2a^2{b}^{-7}{c}^{10}}{6a^{-3}{b}^2{c}^{-3}}$

Answer to Problem 17CYU

The simplified expression is $\frac{a^5{c}^7}{3b^9}$ .

Explanation of Solution

Given information:

  $\frac{2a^2{b}^{-7}{c}^{10}}{6a^{-3}{b}^2{c}^{-3}}$

Concept Used:

Quotient of powers: It states that to divide two powers with same base, subtract the exponent.

For example, let a be any non-zero number and m and n be any integer.

  $\frac{a^m}{a^n}=a^{m-n}$

Negative Exponent Property: $a^{-n}$ is the reciprocal of $a^n$ , where a is a non-zero number and n is an integer.

  $a^{-n}=\frac{1}{a^n}$

Calculation:

  $\frac{2a^2{b}^{-7}{c}^{10}}{6a^{-3}{b}^2{c}^{-3}}$

(Group the powers with the same base)

  $=\left(\frac{2}{6}\right)\left(\frac{a^2}{a^{-3}}\right)\left(\frac{b^{-7}}{b^2}\right)\left(\frac{c^{10}}{c^{-3}}\right)$

Using Quotient of Powers

  $=\left(\frac{2}{6}\right)\left(\frac{a^2}{a^{-3}}\right)\left(\frac{b^{-7}}{b^2}\right)\left(\frac{c^{10}}{c^{-3}}\right)$

Simplify.

  $\begin{array}{l}=\frac{1}{3}{a}^{2-\left(-3\right)}{b}^{-7-2}{c}^{10-3}\\ =\frac{1}{3}{a}^5{b}^{-9}{c}^7\end{array}$

Applying Negative Exponent Property

  $\begin{array}{l}=\frac{1}{3}{a}^5{b}^{-9}{c}^7\\ =\frac{1}{3}{a}^5\frac{1}{b^9}{c}^7\\ =\frac{a^5{c}^7}{3b^9}\end{array}$