Q. 157 - Intermediate Algebra

Question

Simplify 

(a) ${\left(2r^{-2}\right)}^3{\left(4^{-1}r\right)}^2$ (b) ${\left(3x^{-3}\right)}^3{\left(3^{-1}{x}^5\right)}^4$

Step-by-Step Solution

Answer

Simplified expressions are 

(a)$\frac{1}{2r^4}$

(b)$\frac{x^{11}}{3}$

part (a) Step 1 Given information

${\left(2r^{-2}\right)}^3{\left(4^{-1}r\right)}^2$

part (a) Step 2

$$\begin{gathered} {\left(2r^{-2}\right)}^3{\left(4^{-1}r\right)}^2={\left(2\right)}^3{\left(r^{-2}\right)}^3{\left(4^{-1}\right)}^2{\left(r\right)}^2 \\ =8\times r^{-2\times 3}\times 4^{-1\times 2}\times r^2 \\ =8\times r^{-6}\times 4^{-2}\times r^2 \\ =8\times r^{-6+2}\times \frac{1}{16} \\ =\frac{r^{-4}}{2} \\ =\frac{1}{2r^4} \end{gathered}$$

part (b) Step 1 Given information

${\left(3x^{-3}\right)}^3{\left(3^{-1}{x}^5\right)}^4$

part (b) Step 2

$$\begin{gathered} {\left(3x^{-3}\right)}^3{\left(3^{-1}{x}^5\right)}^4={\left(3\right)}^3{\left(x^{-3}\right)}^3{\left(3^{-1}\right)}^4{\left(x^5\right)}^4 \\ =27\times x^{-3\times 3}\times 3^{-1\times 4}\times x^{5\times 4} \\ =27\times x^{-9}\times \frac{1}{3^4}\times x^{20} \\ =27\times x^{-9+20}\times \frac{1}{81} \\ =\frac{x^{11}}{3} \end{gathered}$$