Question

Evaluate:
(i) ${\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}\times {\left(\frac{3}{5}\right)}^0$
(ii) ${\left(\frac{-3}{5}\right)}^{-4}\times {\left(\frac{-2}{5}\right)}^2$
(iii) ${\left(\frac{-2}{3}\right)}^{-3}\times {\left(\frac{-2}{3}\right)}^{-2}$

Answer 1

(i)

​$$\begin{gathered} {\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}\times {\left(\frac{3}{5}\right)}^0={\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3+0} \\ ={\left(\frac{5}{9}\right)}^{-2}\times {\left(\frac{3}{5}\right)}^{-3}={\left(\frac{9}{5}\right)}^2\times {\left(\frac{5}{3}\right)}^3 \\ =\frac{9^2}{5^2}\times \frac{5^3}{3^3} \\ =\frac{{\left(3^2\right)}^2}{5^2}\times \frac{5^3}{3^3} \\ =\frac{3^4}{5^2}\times \frac{5^3}{3^3}=\left(3^{(4-3)}\right)\times \left(5^{(3-2)}\right)=3\times 5=15 \end{gathered}$$


(ii)

​$$\begin{gathered} {\left(\frac{-3}{5}\right)}^{-4}\times {\left(\frac{-2}{5}\right)}^2={\left(\frac{5}{-3}\right)}^4\times {\left(\frac{-2}{5}\right)}^2 \\ =\frac{5^4}{-3^4}\times \frac{-2^2}{5^2}=5^{(4-2)}\times \frac{-2^2}{-3^4}=5^2\times \frac{-2^2}{-3^4} \\ =25\times \frac{4}{81}=\frac{100}{81} \end{gathered}$$


(iii)

$$\begin{gathered} {\left(\frac{-2}{3}\right)}^{-3}\times {\left(\frac{-2}{3}\right)}^{-2}={\left(\frac{3}{-2}\right)}^3\times {\left(\frac{3}{-2}\right)}^2 \\ =\frac{3^3}{-2^3}\times \frac{3^2}{-2^2}=\frac{3^{(3+2)}}{-2^{(3+2)}}=\frac{3^5}{-2^5}=\frac{-243}{32} \end{gathered}$$
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