Question

Simplify:
a) ${\left({\left(\frac{-2}{3}\right)}^{-2}\right)}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}$ [3 marks]

b) $\frac{49\times z^{-3}}{7^{-3}\times 10\times z^{-5}}(z\ne 0)$ [3 marks]

Answer 1

a) Using the laws of exponents,
$(a^m{)}^n=(a{)}^{m\times n},=a^m\times a^n=a^{m+n}and{a}^m÷a^n=a^{m-n}$

$\therefore {\left({\left(\frac{-2}{3}\right)}^{-2}\right)}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}$

$={\left(\frac{2}{3}\right)}^{-2\times 3}\times (3{)}^4\times \frac{1}{3}\times \frac{1}{6}$
[1 Mark]

$={\left(\frac{-2}{3}\right)}^{-6}\times 3^4\times \frac{1}{3}\times \frac{1}{2\times 3}$

$={\left(\frac{3}{-2}\right)}^6\times 3^4\times \frac{1}{3\times 2\times 3}=\frac{(3{)}^6}{(-2{)}^6}\times 3^4\times \frac{1}{2^1\times 3^2}$

[1 Mark]
=$\frac{(3{)}^6\times (3{)}^4}{(-1{)}^6(2{)}^6\times (2{)}^1\times (3{)}^2}$


$=\frac{(3{)}^{6+4}}{(2{)}^{6+1}\times 3^2}=\frac{(3{)}^{10}}{2^7\times 3^2}$

$=\frac{3^{10-2}}{2^7}=\frac{3^8}{2^7}$
[1 Mark]

b) $\frac{49\times z^{-3}}{7^{-3}\times 10\times z^{-6}}=\frac{(7{)}^2\times z^{-3}}{7^{-3}\times 10z^{-5}}[7^2=49]$

$=\frac{(7{)}^{2-(-3)}\times z^{-3-(-5)}}{10}[a^m\times a^n=a^{m+n}]and[a^m÷a^n=a^{m-n}]$
[1 Mark]

$=\frac{(7{)}^{2+3}\times z^{-3+5}}{10}$


$=\frac{(7{)}^5{z}^2}{10}$

$=\frac{7^5}{10}{z}^2$
[1 Mark]