Question

Simplify each of the following and express each as a rational number:

(i) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{1}{5}\right)}^2$
(ii) ${\left(\frac{-2}{3}\right)}^5\times {\left(\frac{-3}{7}\right)}^3$
(iii) ${\left(\frac{-1}{2}\right)}^5\times 2^3\times {\left(\frac{3}{5}\right)}^2$
(iv) ${\left(\frac{2}{3}\right)}^2\times {\left(\frac{-3}{5}\right)}^3\times {\left(\frac{7}{2}\right)}^2$
(v) $\left\{{\left(\frac{-3}{4}\right)}^3-{\left(\frac{-5}{2}\right)}^3\right\}\times 4^2$

Answer 1

(i) ${\left(\frac{3}{2}\right)}^4\times {\left(\frac{1}{5}\right)}^2=\frac{3^4}{2^4}\times \frac{1^2}{5^2}=\frac{81\times 1}{16\times 25}=\frac{81}{400}$

(ii) ${\left(\frac{-2}{3}\right)}^5\times {\left(\frac{-3}{7}\right)}^3=\frac{{\left(-2\right)}^5}{{\left(3\right)}^5}\times \frac{{\left(-3\right)}^3}{{\left(7\right)}^3}$
= $$\begin{gathered} =\frac{{\left(-2\right)}^5}{{\left(7\right)}^3}\times \frac{\left(-1\right){\left(3\right)}^3}{{\left(3\right)}^5} \\ =\frac{-32\times -1\times 3^{3-5}}{343} \\ =\frac{-32\times -1\times 3^{-2}}{343} \\ =\frac{-32\times -1\times 1}{343\times 9} \\ =\frac{32}{3087} \end{gathered}$$ $\left[s\mathrm{ince}{3}^{-2}=\frac{1}{9}\right]$


(iii) ${\left(\frac{-1}{2}\right)}^5\times 2^3\times {\left(\frac{3}{4}\right)}^2=\frac{{\left(-1\right)}^5}{2^5}\times 2^3\times \frac{3^2}{4^2}$
$=\frac{{\left(-1\right)}^5}{2^5}\times 2^3\times \frac{3^2}{(2^2{)}^2}$
$=\frac{-1\times 2^3\times 3^2}{2^5\times 2^4}$
$=\frac{-1\times 2^3\times 3^2}{2^9}$ $=-1\times 2^{3-9}\times 3^2$ = $-9\times 2^{-6}=\frac{-9}{2^6}=\frac{-9}{64}$ $\left[s\mathrm{ince}{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$

(iv) ${\left(\frac{2}{3}\right)}^2\times {\left(\frac{-3}{5}\right)}^3\times {\left(\frac{7}{2}\right)}^2=\frac{2^2}{3^2}\times \frac{{\left(-3\right)}^3}{5^3}\times \frac{7^2}{2^2}$
$\frac{-1\times 3^{3-2}\times 7^2}{5^3}=\frac{-1\times 3^1\times 7^2}{5^3}=\frac{-1\times 3\times 49}{125}=\frac{-147}{125}$

(v) $\left\{{\left(\frac{-3}{4}\right)}^3-{\left(\frac{-5}{2}\right)}^3\right\}\times 4^2=\left\{\left(\frac{-3^3}{4^3}\right)-\left(\frac{-5^3}{2^3}\right)\right\}\times 4^2$
$=\left\{\left(\frac{-27}{64}\right)-\left(\frac{-125}{8}\right)\right\}\times 16$
$=\left\{\frac{-27}{64}+\frac{125}{8}\right\}\times 16$
$=\left(\frac{-27+1000}{64}\right)\times 16$
$=\left(\frac{973}{64}\times 16\right)=\frac{973}{4}$