# Exercise 2.4 (Solutions)

### Question 1

Use law of exponent to simplify.

• (i) $\frac{(243)^{\frac{-2}{3}}(32)^{\frac{-1}{5}}}{\sqrt(196)^{-1}}$
• (ii) $\left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right)$
• (iii) $\left(\frac{x^{-2}y^{-1}z^{-4}}{x^4y^{-3}z^0}\right)^{-3}$
• (iv) $\frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\right)}{\left(9^2n\right)\left(3^3\right)}$

Solution

(i) $$\begin{array}{cl} \begin{array}{cl} \frac{(243)^{\frac{-2}{3}}(32)^{\frac{-1}{5}}}{\sqrt(196)^{-1}} &= \frac{(3^5)^\frac{-2}{3}(2^5)^\frac{-1}{5}}{(14^{2})^{{-1}\times\frac{1}{2}}}\\ &= \frac{3^{\frac{-10}{3}}2^{-1}}{14^{-1}}\\ &= \frac{3^\frac{-10}{3}2^{-1}}{2^{-1}\times7^{-1}}\\ &= \frac{3^{\frac{-10}{3}}}{7^{-1}}\\ &= \frac{7}{3^{\frac{10}{3}}}\\ &= \frac{7}{3^{(9)\times\frac{1}{3}}\times3^\frac{1}{3}}\\ &= \frac{7}{3^3.\sqrt[3]{3}}\\ &= \frac{7}{27.\sqrt[3]{3}} \end{array}\end{array}$$

(ii) $$\begin{array}{cl} \left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right) &= (2)(-8)x^5.y^{-4}.x^{-3}y^2\\ &= -16 x^{5-3}y^{-4+2}\\ &= -16x^2y^{-2}\\ &= \frac{-16x^2}{y^2} \end{array}$$

(iii) $$\begin{array}{cl} \left(\frac{x^{-2}y^{-1}z^{-4}}{x^4y^{-3}z^0}\right)^{-3} &= \left(\frac{y^{-1+3}}{x^{4+2}z^4}\right)^{-3}\\ &= \left(\frac{y^2}{x^6 z^4}\right)^{-3}\\ &= \frac{y^{2\left(-3\right)}}{x^{6\left(-3\right)}z^{4\left(-3\right)}}\\ &=\frac{y^{-6}}{x^{-18} z^{-12}}\\ &=\frac{x^{18} z^{12}}{y^6} \end{array}$$

(iv) $$\begin{array}{cl} \frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\right)}{\left(9^2n\right)\left(3^3\right)} &= \frac{\left(3^4 \right )^n.3^5 -3^{4n-1}\left(3 \right )^5}{\left(3^2 \right )^2n.3^3}\\ &= \frac{3^{4n+5}-3^{4n-1+5}}{3^{4n}.3^3}\\ &= \frac{3^{4n+4}.3^1-3^{4n+4}}{3^{4n+3}}\\ &= \frac{3^{4n+4}\left(3-1 \right )}{3^{4n+3}}\\ &= 3^{4n+4-4n-3}\times2\\ &= 3\times2\\ &= 6 \end{array}$$

### Question 2

Show that $$\left(\frac{x^a}{x^b}\right )^{a+b}\times\left(\frac{x^b}{x^c}\right )^{b+c}\times\left(\frac{x^c}{x^a}\right )^{c+a}= 1$$

Solution
$$\begin{array}{cl} \left(x^{a-b}\right)^{a+b}\times\left(x^{b-c}\right)^{b+c}\times\left(x^{c-a}\right)^{c+a} &= x^{a^2-b^2}\times{x^{b^2-c^2}}\times{x^{c^2-a^2}}\\ &= x^{a^2-b^2+b^2-c^2+c^2-a^2}\\ &= x^0\\ &= 1 \end{array}$$

### Question 3

• Simplify
• (i) $\frac{2^{1/3}\left(27\right)^{1/3}\left(60\right)^{1/2}}{\left(180\right)^{1/2}\left(4\right)^{-1/3} 9^{1/4}}$
• (ii) $\sqrt\frac{{216}^{\frac{2}{3}}{25}^{\frac{1}{2}}}{{.04}^{\frac{-1}{2}}}$
• (iii) ${5^2}^3\div\left(5^2 \right )^3$
• (iv) $\left(x^3 \right )^2\div{x^3}^2$

Soluton

(i) $$\begin{array}{cl} \frac{2^{1/3}\left(27\right)^{1/3}\left(60\right)^{1/2}}{\left(180\right)^{1/2}\left(4\right)^{-1/3}9^{1/4}} &= \frac{2^{1/3}\left(3^3\right)^{1/3}\left(2^2.3.5\right)^{1/2}}{\left(2^2.3^2.5\right)^{1/2}\left(2^2 \right)^{-1/3}\left(3^2\right)^{1/4}}\\ &= \frac{2^{1/3}\left(3^3\right)^{1/3}\left(2^2.3.5\right)^{1/2}}{\left(2^2.3^2.5\right)^{1/2}\left(2^2 \right)^{-1/3}\left(3^2\right)^{1/4}}\\ &= \frac{2^{1/3}3.2.3^{1/2}.5^{1/2}}{2.3.5^{1/2}2^{-2/3}3^{1/2}}\\ &= 2^{{1/3}+{2/3}}\\ &= 2^{1+2/3}\\ &= 2^{3/3}\\ &= 2 \end{array}$$ (ii) $$\begin{array}{cl} \sqrt\frac{{216}^{2/3}{25}^{1/2}}{{.04}^{-1/2}} &= \sqrt\frac{{6^3}^{2/3}{5^2}^{1/2}}{{\frac{4}{100}}^{-1/2}}\\ &= \sqrt\frac{{6^2}5}{\frac{1}{25}^{-1/2}}\\ &= \sqrt\frac{{6^2}.5}{\left(\frac{1}{5^2}\right)^{-1/2}}\\ &= \sqrt\frac{{6^2}.5}{5}\\ &= {6^2}^{1/2}\\ &= 6 \end{array}$$

(iii) $$\begin{array}{cl} {5^2}^3\div\left(5^2\right)^3 &= {5^8}\div5^6\\ &= \frac{5^8}{5^6}\\ &= {5^{8-6}}\\ &= {5^2}\\ &= 25 \end{array}$$

(iv) $$\begin{array}{cl} \left(x^3\right)^2\div{x^3}^2 &= x^6\div{x^9}\\ &= \frac{x^6}{x^9}\\ &= \frac{1}{x^{6-9}}\\ &= \frac{1}{x^{-3}}\\ &= \frac{1}{x^3} \end{array}$$