Question

Assuming that x, y, z are positive real numbers, simplify each of the following:

(i) ${\left(\sqrt{x^{-3}}\right)}^5$

(ii) $\sqrt{x^3{y}^{-2}}$

(iii) $(x^{-2/3}{y}^{-1/2}{)}^2$

(iv) $(\sqrt{x}{)}^{-2/3}\sqrt{y^4}÷\sqrt{x{y}^{-1/2}}$

(v) $5\sqrt{243x^{10}{y}^5{z}^{10}}$

(vi) ${\left(\frac{x^{-4}}{y^{-10}}\right)}^{5/4}$

(vii) ${\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^5{\left(\frac{6}{7}\right)}^2$

Answer 1

(i) ${\left(\sqrt{x^{-3}}\right)}^5={\left(x^{\frac{-3}{2}}\right)}^5=x^{\frac{-3}{2}\times 5}$ $=x^{\frac{-15}{2}}=\frac{1}{x^{\frac{15}{2}}}\left\{\begin{array}{c}\because (x^m{)}^n=x^{mn}\\ x^{-m}=\frac{1}{x^m}\end{array}\right\}$

(ii) $\sqrt{x^3{y}^{-2}}={\left(x^3{y}^{-2}\right)}^{\frac{1}{2}}=x^{\frac{3}{2}}.y^{\frac{-2}{2}}=x^{\frac{3}{2}}.y^{-1}\left\{\begin{array}{c}\because (x^m{)}^n=x^{mn}\\ and{x}^{-m}=\frac{1}{x^m}\end{array}\right\}=\frac{x^{\frac{3}{2}}}{y}$

(iii) ${\left(x^{\frac{-2}{3}}{y}^{\frac{-1}{2}}\right)}^2=x^{\frac{-2}{3}\times 2}.y^{\frac{-1}{2}\times 2}=x^{\frac{-4}{3}}.y^{-1}=\frac{1}{x^{\frac{4}{3}}y}\left\{\begin{array}{c}\because (x^m{)}^n=x^{mn}\\ x^{-m}=\frac{1}{x^m}\end{array}\right\}$

(iv) $(\sqrt{x}{)}^{-2/3}\sqrt{y^4}÷\sqrt{x{y}^{-1/2}}={\left(x^{\frac{1}{2}}\right)}^{\frac{-2}{3}}{\left(y^{\frac{1}{2}}\right)}^4÷x^{\frac{1}{2}}{\left(y^{\frac{-1}{2}}\right)}^{\frac{1}{2}}=x^{\frac{1}{2}\times \left(\frac{-2}{3}\right)}.y^{\frac{1}{2}\times 4}÷x^{\frac{1}{2}}.y^{\frac{-1}{2}\times \frac{1}{2}}=x^{\frac{-1}{3}}{y}^2÷x^{\frac{1}{2}}.y^{\frac{-1}{4}}=x^{\frac{-1}{3}-\left(\frac{1}{2}\right)}{y}^{2-\left(\frac{-1}{4}\right)}=x^{\frac{-2-3}{6}}.y^{2+\frac{1}{4}}=x^{\frac{-5}{6}}.y^{\frac{9}{4}}=\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$

(v) $\sqrt[5]{5243x^{10}{y}^5{z}^{10}}$ = ${\left(243x^{10}{y}^5{z}^{10}\right)}^{\frac{1}{5}}={(3^5.x^{10}.y^5.z^{10})}^{\frac{1}{5}}=3^{5\times \frac{1}{5}}.x^{10\times \frac{1}{5}}.y^{5\times \frac{1}{5}}.z^{10\times \frac{1}{5}}=3x^{2}y{z}^2$

(vi) ${\left(\frac{x^{-4}}{y^{-10}}\right)}^{5/4}$ = ${\left(\frac{y^{10}}{x^4}\right)}^{\frac{5}{4}}(\begin{array}{c}\because x^{-m}=\frac{1}{x^m}\end{array})=\frac{y^{\frac{10\times 5}{4}}}{x^{4\times \frac{5}{4}}}=\frac{y^{\frac{25}{2}}}{x^5}$

(vii) ${\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^5{\left(\frac{6}{7}\right)}^2$ $={\left(\frac{2}{3}\right)}^{\frac{5}{2}}\times \frac{36}{49}={\left(\frac{2}{3}\right)}^2\times \frac{\sqrt{2}}{\sqrt{3}}\times \frac{36}{49}=\frac{16\sqrt{2}}{49\sqrt{3}}=\frac{\sqrt{2\times 16\times 16}}{\sqrt{3\times 49\times 49}}=\frac{\sqrt{512}}{\sqrt{7203}}={\left(\frac{512}{7203}\right)}^{\frac{1}{2}}$