RD Sharma 2020 solution class 9 chapter 2 Exponents of Real Numbers FBQS

FBQS

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Question 1:

(21² – 15²)^2/3 is equal to ________.

Answer 1:

$$\begin{gathered} {\left({21}^2-{15}^2\right)}^{\frac{2}{3}}={\left(441-225\right)}^{\frac{2}{3}} \\ ={\left(216\right)}^{\frac{2}{3}} \\ ={\left(6^3\right)}^{\frac{2}{3}} \\ ={\left(6\right)}^{3\times \frac{2}{3}} \\ ={\left(6\right)}^2 \\ =36 \end{gathered}$$

Hence, (21² – 15²)^2/3 is equal to 36.

Question 2:

81^1/4 × 9^3/2 × 27^–4/3 is equal to _________.

Answer 2:

$$\begin{gathered} {81}^{\frac{1}{4}}\times 9^{\frac{3}{2}}\times {27}^{-\frac{4}{3}}={\left(3^4\right)}^{\frac{1}{4}}\times {\left(3^2\right)}^{\frac{3}{2}}\times \frac{1}{{27}^{{\displaystyle \frac{4}{3}}}} \\ ={\left(3\right)}^{4\times \frac{1}{4}}\times {\left(3\right)}^{2\times \frac{3}{2}}\times \frac{1}{{\left(3^3\right)}^{{\displaystyle \frac{4}{3}}}} \\ ={\left(3\right)}^1\times {\left(3\right)}^3\times \frac{1}{{\left(3\right)}^{3\times {\displaystyle \frac{4}{3}}}} \\ ={\left(3\right)}^{1+3}\times \frac{1}{3^{{\displaystyle 4}}} \\ ={\left(3\right)}^4\times \frac{1}{3^{{\displaystyle 4}}} \\ =1 \end{gathered}$$

Hence, 81^1/4 × 9^3/2 × 27^–4/3 is equal to 1.

Question 3:

${\left\{\frac{{\left(169\right)}^{-3}}{{\left(196\right)}^{-8}}\right\}}^{\frac{1}{48}}=$ __________.

Answer 3:

$$\begin{gathered} {\left\{\frac{{\left(169\right)}^{-3}}{{\left(196\right)}^{-8}}\right\}}^{\frac{1}{48}}={\left\{\frac{{\left({13}^2\right)}^{-3}}{{\left({14}^2\right)}^{-8}}\right\}}^{\frac{1}{48}} \\ ={\left\{\frac{{\left(13\right)}^{2\times \left(-3\right)}}{{\left(14\right)}^{2\times \left(-8\right)}}\right\}}^{\frac{1}{48}} \\ ={\left\{\frac{{\left(13\right)}^{-6}}{{\left(14\right)}^{-16}}\right\}}^{\frac{1}{48}} \\ =\frac{{\left(13\right)}^{-6\times {\displaystyle \frac{1}{48}}}}{{\left(14\right)}^{-16\times {\displaystyle \frac{1}{48}}}} \\ =\frac{{\left(13\right)}^{-{\displaystyle \frac{1}{8}}}}{{\left(14\right)}^{-{\displaystyle \frac{1}{3}}}} \\ =\frac{{\displaystyle \frac{1}{{13}^{{\displaystyle \frac{1}{8}}}}}}{{\displaystyle \frac{1}{{14}^{{\displaystyle \frac{1}{3}}}}}} \\ =\frac{{\left(14\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(13\right)}^{{\displaystyle \frac{1}{8}}}} \end{gathered}$$

Hence, ${\left\{\frac{{\left(169\right)}^{-3}}{{\left(196\right)}^{-8}}\right\}}^{\frac{1}{48}}=$ $\underline{ \frac{{\left(14\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(13\right)}^{{\displaystyle \frac{1}{8}}}} }$.

Question 4:

If x = 8^2/3 × 32^–2/5, then x^–5 = ________.

Answer 4:

$$\begin{gathered} \mathrm{Let} x=8^{\frac{2}{3}}\times {32}^{-\frac{2}{5}} \\ \Rightarrow x={\left(2^3\right)}^{\frac{2}{3}}\times {\left(2^5\right)}^{-\frac{2}{5}} \\ \Rightarrow x={\left(2\right)}^{3\times \frac{2}{3}}\times {\left(2\right)}^{5\times \left(-\frac{2}{5}\right)} \\ \Rightarrow x={\left(2\right)}^2\times {\left(2\right)}^{-2} \\ \Rightarrow x=2^2\times \frac{1}{2^2} \\ \Rightarrow x=1 \\ \mathrm{Now}, \\ {x}^{-5}=\frac{1}{x^5} \\ =\frac{1}{1^5} \\ =1 \end{gathered}$$

Hence, if x = 8^2/3 × 32^–2/5, then x^–5 = 1.

Question 5:

If 6ⁿ = 1296, then 6^n–3 = _________.

Answer 5:

$$\begin{gathered} \mathrm{Let} 6^n=1296 \\ \Rightarrow 6^n=6^4 \\ \Rightarrow n=4 \\ \mathrm{Now}, \\ {6}^{n-3}=6^{4-3} \\ =6^1 \\ =6 \end{gathered}$$

Hence, if 6ⁿ = 1296, then 6^n–3 = 6.

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Question 6:

The value of 4 × (256)^–1/4 ÷ (243)^1/5 is ________.

Answer 6:

$$\begin{gathered} \mathrm{Let} x=4\times {\left(256\right)}^{-\frac{1}{4}}÷{\left(243\right)}^{\frac{1}{5}} \\ \Rightarrow x=4\times {\left(4^4\right)}^{-\frac{1}{4}}÷{\left(3^5\right)}^{\frac{1}{5}} \\ \Rightarrow x=4\times {\left(4\right)}^{4\times \left(-\frac{1}{4}\right)}÷{\left(3\right)}^{5\times \frac{1}{5}} \\ \Rightarrow x=4\times {\left(4\right)}^{-1}÷{\left(3\right)}^1 \\ \Rightarrow x=4\times \frac{1}{4}÷{\left(3\right)}^1 \\ \Rightarrow x=1÷3 \\ \Rightarrow x=\frac{1}{3} \end{gathered}$$

Hence, the value of 4 × (256)^–1/4 ÷ (243)^1/5 is $\underline{\frac{1}{3}}$.

Question 7:

If ${\left(6x\right)}^6=6^{2^3}, \mathrm{then} x = \_\_\_\_\_\_\_\_.$

Answer 7:

$$\begin{gathered} \mathrm{Let} {\left(6x\right)}^6=6^{2^3} \\ \Rightarrow {\left(6x\right)}^6=6^8 \\ \Rightarrow 6^6{x}^6=6^8 \\ \Rightarrow x^6=\frac{6^8}{6^6} \\ \Rightarrow x^6=6^{8-6} \\ \Rightarrow x^6=6^2 \\ \Rightarrow x={\left(6^2\right)}^{\frac{1}{6}} \\ \Rightarrow x=6^{2\times \frac{1}{6}} \\ \Rightarrow x=6^{\frac{1}{3}} \end{gathered}$$

Hence, if ${\left(6x\right)}^6=6^{2^3}, \mathrm{then} x=\underline{6^{\frac{1}{3}}}.$

Question 8:

${\left\{{\left(\frac{a}{b}\right)}^{\sqrt{99}-\sqrt{97}}\right\}}^{\sqrt{99}+\sqrt{97}}$= ___________.

Answer 8:

$$\begin{gathered} {\left\{{\left(\frac{a}{b}\right)}^{\sqrt{99}-\sqrt{97}}\right\}}^{\sqrt{99}+\sqrt{97}} \\ ={\left(\frac{a}{b}\right)}^{\left(\sqrt{99}-\sqrt{97}\right)\times \left(\sqrt{99}+\sqrt{97}\right)} \\ ={\left(\frac{a}{b}\right)}^{{\left(\sqrt{99}\right)}^2-{\left(\sqrt{97}\right)}^2} \left(\mathrm{using} \mathrm{the} \mathrm{identity}: a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ ={\left(\frac{a}{b}\right)}^{99-97} \\ ={\left(\frac{a}{b}\right)}^2 \\ =\frac{a^2}{b^2} \end{gathered}$$

Hence, ${\left\{{\left(\frac{a}{b}\right)}^{\sqrt{99}-\sqrt{97}}\right\}}^{\sqrt{99}+\sqrt{97}}$= $\underline{\frac{a^2}{b^2}}.$

Question 9:

If $\frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x, \mathrm{then} x =\_\_\_\_\_\_\_\_\_.$

Answer 9:

$$\begin{gathered} \frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left({\displaystyle \frac{qp+1}{q}}\right)}^{p-q}{\left({\displaystyle \frac{pq-1}{q}}\right)}^{p+q}}{{\left({\displaystyle \frac{pq+1}{p}}\right)}^{p-q}{\left({\displaystyle \frac{pq-1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\displaystyle \frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}}\times {\displaystyle \frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}}}{{\displaystyle \frac{{\left(pq+1\right)}^{p-q}}{{\left(p\right)}^{p-q}}}\times {\displaystyle \frac{{\left(pq-1\right)}^{p+q}}{{\left(p\right)}^{p+q}}}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}\times \frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}\times \frac{{\left(p\right)}^{p-q}}{{\left(pq+1\right)}^{p-q}}\times \frac{{\left(p\right)}^{p+q}}{{\left(pq-1\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{p-q}}{{\left(q\right)}^{p-q}}\times \frac{{\left(p\right)}^{p+q}}{{\left(q\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{p-q+p+q}}{{\left(q\right)}^{p-q+p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{2p}}{{\left(q\right)}^{2p}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow {\left(\frac{p}{q}\right)}^{2p}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow 2p=x \\ \Rightarrow x=2p \end{gathered}$$

Hence, if $\frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x, \mathrm{then} x=\underline{2p}.$

Question 10:

If 5ⁿ⁺² = 625, then (12n + 3)^1/3 = _________.

Answer 10:

$$\begin{gathered} \mathrm{Let} 5^{n+2}=625 \\ \Rightarrow 5^{n+2}=5^4 \\ \Rightarrow n+2=4 \\ \Rightarrow n=2 \\ \mathrm{Now}, \\ {\left(12n+3\right)}^{\frac{1}{3}}={\left(12\left(2\right)+3\right)}^{\frac{1}{3}} \\ ={\left(24+3\right)}^{\frac{1}{3}} \\ ={\left(27\right)}^{\frac{1}{3}} \\ ={\left(3^3\right)}^{\frac{1}{3}} \\ ={\left(3\right)}^{3\times \frac{1}{3}} \\ =3 \end{gathered}$$

Hence, if 5ⁿ⁺² = 625, then (12n + 3)^1/3 = 3.

Question 11:

If ${\left(\frac{a^{3-7a}\times a^{6-2a}}{a^{2a}\times a^{9-2a}}\right)}^{\frac{1}{9}}=$__________.

Answer 11:

$$\begin{gathered} {\left(\frac{a^{3-7a}\times a^{6-2a}}{a^{2a}\times a^{9-2a}}\right)}^{\frac{1}{9}}={\left(\frac{a^{3-7a+6-2a}}{a^{2a+9-2a}}\right)}^{\frac{1}{9}} \\ ={\left(\frac{a^{9-9a}}{a^9}\right)}^{\frac{1}{9}} \\ ={\left(a^{9-9a-9}\right)}^{\frac{1}{9}} \\ ={\left(a^{-9a}\right)}^{\frac{1}{9}} \\ ={\left(a\right)}^{-9a\times \frac{1}{9}} \\ ={\left(a\right)}^{-a} \\ =\frac{1}{a^a} \end{gathered}$$

Hence, ${\left(\frac{a^{3-7a}\times a^{6-2a}}{a^{2a}\times a^{9-2a}}\right)}^{\frac{1}{9}}=\underline{\frac{1}{a^a}}.$ 

Question 12:

If $\frac{1}{{\left(243\right)}^x}={\left(729\right)}^y=3^3$, then 5x + 6y = __________.

Answer 12:

$$\begin{gathered} \mathrm{Given}: \frac{1}{{\left(243\right)}^x}={\left(729\right)}^y=3^3 \\ \frac{1}{{\left(243\right)}^x}=3^3 \\ \Rightarrow \frac{1}{{\left(3^5\right)}^x}=3^3 \\ \Rightarrow \frac{1}{3^{5x}}=3^3 \\ \Rightarrow 3^{-5x}=3^3 \\ \Rightarrow -5x=3 \\ \Rightarrow 5x=-3 ...\left(1\right) \\ {\left(729\right)}^y=3^3 \\ \Rightarrow {\left(3^6\right)}^y=3^3 \\ \Rightarrow 3^{6y}=3^3 \\ \Rightarrow 6y=3 ...\left(2\right) \\ \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ 5x+6y=-3+3 \\ =0 \end{gathered}$$

Hence, 5x + 6y = 0.

Question 13:

If 6^x–y = 36 and 3^x+y = 729, then x² – y² = _________.

Answer 13:

$$\begin{gathered} \mathrm{Given}: 6^{x-y}=36 \\ \mathrm{and} 3^{x+y}=729 \\ 6^{x-y}=36 \\ \Rightarrow 6^{x-y}=6^2 \\ \Rightarrow x-y=2 ...\left(1\right) \\ {3}^{x+y}=729 \\ \Rightarrow 3^{x+y}=3^6 \\ \Rightarrow x+y=6 ...\left(2\right) \\ \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ 2x=8 \\ \Rightarrow x=4 \\ \mathrm{Substituting} \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{in} \left(2\right), \mathrm{we} \mathrm{get} \\ 4+y=6 \\ \Rightarrow y=2 \\ \mathrm{Now}, \\ {x}^2-y^2=4^2-2^2 \\ =16-4 \\ =12 \end{gathered}$$

Hence, x² – y² = 12.

Question 14:

$\sqrt[4]{\sqrt[3]{2^2}}$ equals __________.

Answer 14:

$$\begin{gathered} \sqrt[4]{\sqrt[3]{2^2}}={\left\{{\left(2^2\right)}^{\frac{1}{3}}\right\}}^{\frac{1}{4}} \\ ={\left(2^2\right)}^{\frac{1}{3}\times \frac{1}{4}} \\ ={\left(2^2\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{2\times \frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{6}} \end{gathered}$$

Hence, $\sqrt[4]{\sqrt[3]{2^2}}$ equals $\underline{{\left(2\right)}^{\frac{1}{6}}}.$

Question 15:

The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ is equal to ________.

Answer 15:

$$\begin{gathered} \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(32\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(2^5\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(2\right)}^{5\times \frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}.{\left(2\right)}^{\frac{1}{4}}.{\left(2\right)}^{\frac{5}{12}} \\ ={\left(2\right)}^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}} \\ ={\left(2\right)}^{\frac{4+3+5}{12}} \\ ={\left(2\right)}^{\frac{12}{12}} \\ ={\left(2\right)}^1 \\ =2 \end{gathered}$$

Hence, the product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ is equal to 2.

Question 16:

$\sqrt[4]{{\left(81\right)}^{-2}}$ is equal to _________.

Answer 16:

$$\begin{gathered} \sqrt[4]{{\left(81\right)}^{-2}}={\left\{{\left(81\right)}^{-2}\right\}}^{\frac{1}{4}} \\ ={\left(81\right)}^{-2\times \frac{1}{4}} \\ ={\left(81\right)}^{-\frac{1}{2}} \\ ={\left(3^4\right)}^{-\frac{1}{2}} \\ ={\left(3\right)}^{4\times \left(-\frac{1}{2}\right)} \\ ={\left(3\right)}^{-2} \\ =\frac{1}{3^2} \\ =\frac{1}{9} \end{gathered}$$

Hence, $\sqrt[4]{{\left(81\right)}^{-2}}$ is equal to $\underline{\frac{1}{9}}$.

Question 17:

The value of (256)^0.16  × (256)^0.09 is ________.

Answer 17:

$$\begin{gathered} {\left(256\right)}^{0.16}\times {\left(256\right)}^{0.09}={\left(256\right)}^{0.16+0.09} \\ ={\left(256\right)}^{0.25} \\ ={\left(256\right)}^{\frac{25}{100}} \\ ={\left(256\right)}^{\frac{1}{4}} \\ ={\left(2^8\right)}^{\frac{1}{4}} \\ ={\left(2\right)}^{8\times \frac{1}{4}} \\ ={\left(2\right)}^2 \\ =4 \end{gathered}$$

Hence, the value of (256)^0.16  × (256)^0.09 is 4.