Chapter 2.1, Problem 7GP

Solution

To simplify the given expression ${\left(\frac{s^3}{t^{-4}}\right)}^2$ and tell property used.

The value of ${\left(\frac{s^3}{t^{-4}}\right)}^2$ is $s^6{t}^8$ and property used is power of quotient ${\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}$ , $b\ne 0$ , power of power ${\left(a^n\right)}^m=a^{nm}$ and negative exponent $a^{-m}=\frac{1}{a^m},a\ne 0$ .

Given:

Expression: ${\left(\frac{s^3}{t^{-4}}\right)}^2$ .

Concept Used:

Property is used power of quotient ${\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}$ , $b\ne 0$ , power of power ${\left(a^n\right)}^m=a^{nm}$ and negative exponent $a^{-m}=\frac{1}{a^m},a\ne 0$ .

Calculation:

The power of quotient property and power of power ${\left(a^n\right)}^m=a^{nm}$ are used to solve the question.

  $\begin{array}{c}{\left(\frac{s^3}{t^{-4}}\right)}^2=\frac{s^6}{t^{-8}}\text{Such that }t\ne 0\\ =s^6{t}^{-\left(-8\right)}\\ =s^6{t}^8\end{array}$

Thus, final value is $\frac{s^6}{t^{-8}}=s^6{t}^8$ .