Question

Match the following columns:

Column I		Column II
(a)	$\sqrt[4]{{\left(81\right)}^{-2}}=.......$	(p)	4
(b)	If ${\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{b}{a}\right)}^{x-4}$, then x = ........ .	(q)	$\frac{2}{9}$
(c)	If $x=\left(9+4\sqrt{5}\right)$, then $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$= ...... .	(r)	$\frac{1}{9}$
(d)	${\left(\frac{81}{16}\right)}^{-3/4}\times {\left(\frac{64}{27}\right)}^{-1/3}=?$	(s)	3

(a) ......
(b) ......
(c) ......
(d) ......

Answer 1

(a)
$$\begin{gathered} {\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}} \\ ={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4\times \frac{1}{4}}={\left(9\right)}^{-1} \\ =\frac{1}{9} \end{gathered}$$

(b)

$$\begin{gathered} {\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{b}{a}\right)}^{x-4} \\ \Rightarrow {\left(\frac{b}{a}\right)}^{2-x}={\left(\frac{b}{a}\right)}^{x-4} \\ \Rightarrow 2-x=x-4 \\ \Rightarrow 2x=6 \\ \Rightarrow x=3 \end{gathered}$$

(c)

$$\begin{gathered} x=9+4\sqrt{5} \\ \mathrm{and} \\ \frac{1}{x}=\frac{1}{9+4\sqrt{5}}\times \frac{9-4\sqrt{5}}{9-4\sqrt{5}} \\ =\frac{9-4\sqrt{5}}{81-80} \\ =9-4\sqrt{5} \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ x+\frac{1}{x}=9+4\sqrt{5}+9-4\sqrt{5}=18 \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ x+\frac{1}{x}=18 \\ \mathrm{We}\mathrm{know}: \\ {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=\mathrm{x}+\frac{1}{\mathrm{x}}-2\times \mathrm{x}\times \frac{1}{\mathrm{x}} \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=18-2 \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=16 \\ \mathrm{Taking}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}=4 \end{gathered}$$

(d)

$$\begin{gathered} {\left(\frac{3}{2}\right)}^{4\times \frac{-3}{4}}\times {\left(\frac{4}{3}\right)}^{3\times \frac{-1}{3}} \\ ={\left(\frac{3}{2}\right)}^{-3}\times {\left(\frac{4}{3}\right)}^{-1} \\ ={\left({\displaystyle \frac{3}{2}}\right)}^{-3}\times \frac{3}{4} \\ =\left({\displaystyle \frac{3^{-3}}{2^{-3}}}\right)\times \frac{3}{2^2} \\ =\left({\displaystyle \frac{3^{-3}\times 3}{2^{-3}\times 2^2}}\right) \\ =\frac{3^{-2}}{2^{-1}} \\ =\frac{2}{9} \end{gathered}$$