Q14. - Algebra 1

Question

Simplify the expression. Show the work

${\left(\frac{{\displaystyle -2r^{-2}{q}^5{t}^2}}{{\displaystyle 5r^4{q}^2{t}^{-3}}}\right)}^{\mathbf{-}\mathbf{2}}$

Step-by-Step Solution

Answer

The simplified version of ${\left(\frac{{\displaystyle -2r^{-2}{q}^5{t}^2}}{{\displaystyle 5r^4{q}^2{t}^{-3}}}\right)}^{-2}$is $\frac{\mathbf{25}{\mathbf{r}}^{\mathbf{12}}}{\mathbf{4}{\mathbf{q}}^{\mathbf{6}}{\mathbf{t}}^{\mathbf{10}}}$.

Step 1. State the division rule of ‘laws of indices’.

If a fraction is raised to a power, then every terms of the fraction is raised to the power.

In general: ${\left(\frac{{\displaystyle x}}{{\displaystyle y}}\right)}^m=\frac{x^m}{y^m}$

Step 2. State the division rule of ‘laws of indices’.

If the two terms having the same base are divided, then the base is written only once and the indices are subtracted.

In general: $\frac{x^m}{x^n}=x^{m-n}$

Step 3. State the ‘Law of indices’ for power of index numbers.

If a term with a power is itself raised to a power, then the powers are multiplied together.

In general: ${\left(x^m\right)}^n=x^{m\times n}$

Step 4. State the law of indices for negative index.

If a term has negative index, then the reciprocal of that term is taken and raise it with the positive index.

In general: $x^{-n}=\frac{1}{x^n}$

Step 4. Simplify the expression.

Using laws of indices simplify ${\left(\frac{{\displaystyle -2r^{-2}{q}^5{t}^2}}{{\displaystyle 5r^4{q}^2{t}^{-3}}}\right)}^{-2}$

 $\begin{array}{l}{\left(\frac{-2r^{-2}{q}^5{t}^2}{5r^4{q}^2{t}^{-3}}\right)}^{-2}={\left[\left(\frac{-2}{5}\right)\left(\frac{r{}^{-2}}{r^4}\right)\left(\frac{q^5}{q^2}\right)\left(\frac{t^2}{t^{-3}}\right)\right]}^{-2} \left[\text{Collect\hspace{0.17em}\hspace{0.17em}the\hspace{0.17em}\hspace{0.17em}like\hspace{0.17em}\hspace{0.17em}terms\hspace{0.17em}\hspace{0.17em}together}\right]\\ ={\left[\left(\frac{-2}{5}\right)\left(r^{-2-4}\right)\left(q^{5-2}\right)\left(t^{2-(-3)}\right)\right]}^{-2} [\text{Using }\frac{x^m}{x^n}=x^{m-n}]\\ ={\left[\left(\frac{-2}{5}\right)\left(r^{-6}\right)\left(q^3\right)\left(t^{2+3}\right)\right]}^{-2}\\ ={\left[\left(\frac{-2}{5}\right)\left(r^{-6}\right)\left(q^3\right)\left(t^5\right)\right]}^{-2}\\ ={\left(\frac{-2}{5}\right)}^{-2}{\left(r^{-6}\right)}^{-2}{\left(q^3\right)}^{-2}{\left(t^5\right)}^{-2}[\text{Using\hspace{0.17em}\hspace{0.17em}}{\left(xy\right)}^m=x^m{y}^m]\\ ={\left(\frac{-2}{5}\right)}^{-2}\left(r^{(-6)(-2)}\right)\left(q^{\left(3\right)(-2)}\right)\left(t^{\left(5\right)(-2)}\right) [\text{Using\hspace{0.17em}\hspace{0.17em}}{\left(x^m\right)}^n=x^{m\times n}]\\ ={\left(\frac{-2}{5}\right)}^{-2}\left(r^{12}\right)\left(q^{-6}\right)\left(t^{-10}\right)\\ ={\left(\frac{5}{-2}\right)}^2\left(r^{12}\right)\left(\frac{1}{q^6}\right)\left(\frac{1}{t^{10}}\right) [\text{Using\hspace{0.17em}\hspace{0.17em}}{x}^{-n}=\frac{1}{x^n}]\\ =\frac{{\left(5\right)}^2}{{(-2)}^2}\times \frac{r^{12}}{q^6{t}^{10}} [\text{Using\hspace{0.17em}\hspace{0.17em}}{\left(\frac{x}{y}\right)}^m=\frac{x^m}{y^m}]\\ =\frac{25}{4}\times \frac{r^{12}}{q^6{t}^{10}}\\ =\frac{25r^{12}}{4q^6{t}^{10}}\end{array}$