Question

Simplify

(i) ${\left[{\left(\frac{15}{12}\right)}^3\right]}^4$ (ii) ${\left(3^4\right)}^{-2}$ (iii) ${\left({\left(\frac{1}{7}\right)}^{-3}\right)}^4$ (iv) ${\left({\left(\frac{2}{5}\right)}^{-2}\right)}^{-3}$ (v) ${\left(6^5\right)}^4$

(vi) ${\left[{\left(\frac{6}{7}\right)}^5\right]}^2$ (vii) ${\left[{\left(\frac{2}{3}\right)}^{-4}\right]}^5$ (viii) ${\left[{\left(\frac{5}{8}\right)}^3\right]}^{-2}$ (ix) ${\left[{\left(\frac{3}{4}\right)}^6\right]}^1$ (x) ${\left[{\left(\frac{2}{5}\right)}^{-3}\right]}^2$

Answer 1

It is known that, (a^m)ⁿ = a^mn, where m and n are integers and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right){\left[{\left(\frac{15}{12}\right)}^3\right]}^4 \\ ={\left(\frac{15}{12}\right)}^{3\times 4} \\ ={\left(\frac{15}{12}\right)}^{12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right){\left(3^4\right)}^{-2} \\ =3^{4\times \left(-2\right)} \\ =3^{-8} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right){\left[{\left(\frac{1}{7}\right)}^{-3}\right]}^4 \\ ={\left(\frac{1}{7}\right)}^{-3\times 4} \\ ={\left(\frac{1}{7}\right)}^{-12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right){\left[{\left(\frac{2}{5}\right)}^{-2}\right]}^{-3} \\ ={\left(\frac{2}{5}\right)}^{\left(-2\right)\times \left(-3\right)} \\ ={\left(\frac{2}{5}\right)}^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right){\left(6^5\right)}^4 \\ =6^{5\times 4} \\ =6^{20} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right){\left[{\left(\frac{6}{7}\right)}^5\right]}^2 \\ ={\left(\frac{6}{7}\right)}^{5\times 2} \\ ={\left(\frac{6}{7}\right)}^{10} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vii}\right){\left[{\left(\frac{2}{3}\right)}^{-4}\right]}^5 \\ ={\left(\frac{2}{3}\right)}^{\left(-4\right)\times 5} \\ ={\left(\frac{2}{3}\right)}^{-20} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{viii}\right){\left[{\left(\frac{5}{8}\right)}^3\right]}^{-2} \\ ={\left(\frac{5}{8}\right)}^{3\times \left(-2\right)} \\ ={\left(\frac{5}{8}\right)}^{-6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ix}\right){\left[{\left(\frac{3}{4}\right)}^6\right]}^1 \\ ={\left(\frac{3}{4}\right)}^{6\times 1} \\ ={\left(\frac{3}{4}\right)}^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{x}\right){\left[{\left(\frac{2}{5}\right)}^{-3}\right]}^2 \\ ={\left(\frac{2}{5}\right)}^{\left(-3\right)\times 2} \\ ={\left(\frac{2}{5}\right)}^{-6} \end{gathered}$$