Question

Find the value of each of the following:
(i) ${\left(\frac{-1}{2}\right)}^2\times 2^3\times {\left(\frac{3}{4}\right)}^2$
(ii) ${\left(\frac{-3}{5}\right)}^4\times {\left(\frac{4}{9}\right)}^4\times {\left(\frac{-15}{18}\right)}^2$

Answer 1

We have

(i)
${\left(\frac{-1}{2}\right)}^2\times 2^3\times {\left(\frac{3}{4}\right)}^2=\frac{1}{4}\times 8\times \frac{9}{16}=\frac{1}{4}\times \frac{9}{2}=\frac{9}{8}\left[\mathrm{Since}{\left(\frac{-1}{2}\right)}^2=\frac{1}{4},2^3=8\mathrm{and}{\left(\frac{3}{4}\right)}^2=\frac{9}{16}\right]$

(ii)
$$\begin{gathered} {\left(\frac{-3}{5}\right)}^4\times {\left(\frac{4}{9}\right)}^4\times {\left(\frac{-15}{18}\right)}^2 \\ =\frac{(-3{)}^4}{5^4}\times \frac{4^4}{9^4}\times {\left(\frac{-3\times 5}{2\times 9}\right)}^2 \\ =\frac{(-3{)}^4}{5^4}\times \frac{4^4}{9^2\times 9^2}\times {\left(\frac{-3\times 5}{2\times 9}\right)}^2 \\ =\frac{81}{5^4}\times \frac{4^4}{81\times 9^2}\times \frac{(-3{)}^2\times 5^2}{2^2\times 9^2} \\ =\frac{1}{5^4}\times \frac{4^4}{9^2}\times \frac{(-3{)}^2\times 5^2}{2^2\times 9^2} \\ =\frac{1}{5^4}\times \frac{4^4}{9^2}\times \frac{9\times 5^2}{4\times 9^2} \\ =\frac{1}{5^2}\times \frac{4^3}{9^2}\times \frac{1}{9} \\ =\frac{1\times 4^3}{5^2\times 9^3}=\frac{64}{25\times 729}=\frac{64}{18225}\left[\mathrm{Since}{4}^3=64,5^2=25\mathrm{and}{9}^3=729\right] \end{gathered}$$