Chapter 12, Problem 16CLR

To determine

To Evaluate: The simplified value of the expression $\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}$ .

Answer to Problem 16CLR

The value of the expression $\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}$ is $\frac{1}{3r}$ .

Explanation of Solution

Given information:

The expression is $\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}$ .

Formula used:

To find a power of a power, multiply the exponents. That is,

  ${\left(a^m\right)}^n=a^{mn}$

When an algebraic term is divided by another algebraic term power of same variable subtracted and the numerical coefficients divided.

Example-:

  $\frac{7a^8}{14a^2}=\frac{7}{14}{a}^{8-2}$

Calculation:

Consider the expression is $\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}$ .

Recall that to find a power of a power, multiply the exponents. That is,

  ${\left(a^m\right)}^n=a^{mn}$

Use it,

  $\begin{array}{c}\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}=\frac{3^3{r}^{3\times 3}{s}^{4\times 3}}{{\left(-9\right)}^2{r}^{5\times 2}{s}^{6\times 2}}\\ =\frac{27r^9{s}^{12}}{81r^{10}{s}^{12}}\end{array}$

Again recall that when an algebraic term is divided by another algebraic term power of same variable subtracted and the numerical coefficients divided.

Apply it,

  $\begin{array}{c}\frac{27r^9{s}^{12}}{81r^{10}{s}^{12}}=\frac{27}{81}{r}^{9-10}{s}^{12-12}\\ =\frac{1}{3}{r}^{-1}{s}^0\\ =\frac{1}{3r}\end{array}$

Thus, the value of the expression $\frac{{\left(3r^3{s}^4\right)}^3}{{\left(-9r^5{s}^6\right)}^2}$ is $\frac{1}{3r}$ .