Chapter 8: Exponents and Powers

Ans: Given,

Sun’s diameter = 1.4 × 109 m

Earth’s diameter = 1.2756 × 107 m

For comparing, we need to change the diameters of both sun

and earth in the same powers of 10, i.e. 1.2756 x 107 = 0.012756 x 109

Now, if the diameter of sun is divided by the diameter of the earth = $\dfrac{{1.4 \times {{10}^9}}}{{0.012756 \times {{10}^9}}} = 110$

Therefore, the sun’s diameter is 110 times the earth’s diameter.

Ans: Mass of Mars = 6.42 x 1029 kg

Mass of the Sun = 1.99 x 1030 kg 

Total mass of Mars and Sun together = 6.42 x 1029 + 1.99 x 1030 

= 6.42 x 1029 + 19.9 x 1029 = 26.32 x 1029 kg

Ans: The distance between the Sun and the Earth is 1.496 x 108 km

= 1.496 x 108 x 103 m = 1496 x 108 m

The distance between the Earth and the Moon is 3.84 x108 m.

The distance between the Moon and the Sun at particular time (solar eclipse) = (1496 x 108 – 3.84 x 108)m = 1492.16 x 108 m

Ans: Total distance between the earth and the star = 8.1 x 1013 km = 8.1 x 1013 x 103 m

Speed of light = 3 x 108 m per second.

Hence, time that light would take to reach the earth = $\dfrac{{8.1 \times {{10}^{13}} \times {{10}^3}}}{{3 \times {{10}^8}}}$

= $\dfrac{{8.1 \times {{10}^{16}}}}{{3 \times {{10}^8}}}$

= $\dfrac{{8.1}}{3} \times {10^8}$

= $2.7 \times {10^8}s$ 

Ans: Let the number be x.

Hence,

${\left( { - 15} \right)^{ - 1}} \div x = {\left( { - 5} \right)^{ - 1}}$

$\Rightarrow \dfrac{1}{{ - 15}} \times \dfrac{1}{x} = \dfrac{1}{{ - 5}}$

$\Rightarrow x = \dfrac{{ - 15}}{{ - 5}} = 3$

Therefore, the required number is 3.

Ans: Let the number be x.

Hence,

$x \times {\left( { - 8} \right)^{ - 3}} = {\left( { - 6} \right)^{ - 3}}$

$\Rightarrow x = \dfrac{{{{\left( { - 6} \right)}^{ - 3}}}}{{{{\left( { - 8} \right)}^{ - 3}}}} = \dfrac{{{{\left( { - 8} \right)}^3}}}{{{{\left( { - 6} \right)}^3}}} = \dfrac{{ - {8^3}}}{{ - {6^3}}}$

= $\dfrac{{512}}{{216}} = \dfrac{{64}}{{27}}$

Therefore, the required number is $\dfrac{{64}}{{27}}$.

Ans: Use law of exponents: ${a^m} \div {a^n} = {a^{m - n}}$,

${\left( { - \dfrac{1}{7}} \right)^{ - 5}} \div {\left( { - \dfrac{1}{7}} \right)^{ - 7}} = {\left( { - 7} \right)^x}$

$\Rightarrow {\left( { - \dfrac{1}{7}} \right)^{ - 5 - \left( { - 7} \right)}} = {\left( { - 7} \right)^x}$

$\Rightarrow {\left( { - \dfrac{1}{7}} \right)^{ - 5 + 7}} = {\left( { - 7} \right)^x}$

$\Rightarrow {\left( { - \dfrac{1}{7}} \right)^2} = {\left( { - 7} \right)^x}$

$\Rightarrow {\left( { - 7} \right)^{ - 2}} = {\left( { - 7} \right)^x}$

Compare both sides,

x = -2

(2) ${\left( {\dfrac{2}{5}} \right)^{2x + 6}} \times {\left( {\dfrac{2}{5}} \right)^3} = {\left( {\dfrac{2}{5}} \right)^{x + 2}}$

Ans: Use law of exponents: ${a^m} \times {a^n} = {a^{m + n}}$,

${\left( {\dfrac{2}{5}} \right)^{2x + 6}} \times {\left( {\dfrac{2}{5}} \right)^3} = {\left( {\dfrac{2}{5}} \right)^{x + 2}}$

$\Rightarrow {\left( {\dfrac{2}{5}} \right)^{2x + 6 + 3}} = {\left( {\dfrac{2}{5}} \right)^{x + 2}}$

$\Rightarrow {\left( {\dfrac{2}{5}} \right)^{2x + 9}} = {\left( {\dfrac{2}{5}} \right)^{x + 2}}$

Compare both sides,

2x + 9 = x + 2

x = 2 – 9 = -7 

(3) ${2^x} + {2^x} + {2^x} = 192$

Ans: ${2^x}\left( {1 + 1 + 1} \right) = 192$

$\Rightarrow {2^x} \times 3 = 192$

$\Rightarrow {2^x} = \dfrac{{192}}{3} \Rightarrow {2^x} = 64$

$\Rightarrow {2^x} = {2^6}$

Compare both sides,

x = 6

(4) ${\left( {\dfrac{{ - 6}}{7}} \right)^{x - 7}} = 1$

Ans: Using law of exponents ${x^0} = 1$

So, ${\left( {\dfrac{{ - 6}}{7}} \right)^{x - 7}} = 1$ is possible if x – 7 = 0

Thus, x = 7

(5) ${2^{3x}} = {8^{2x + 1}}$

Ans:  ${2^{3x}} = {8^{2x + 1}}$

$\Rightarrow {2^{3x}} = {\left( {{2^3}} \right)^{2x + 1}}$

$\Rightarrow {2^{3x}} = {2^{3\left( {2x + 1} \right)}}$

$\Rightarrow {2^{3x}} = {2^{6x + 3}}$

Compare both sides,

$\Rightarrow x =  - 1$

(6) ${5^x} + {5^{x - 1}} = 750$

Ans: ${5^x} + {5^{x - 1}} = 750$

$\Rightarrow {5^x} + {5^x} \cdot {5^{ - 1}} = 750$

$\Rightarrow {5^x}\left( {1 + {5^{ - 1}}} \right) = 750$

$\Rightarrow {5^x}\left( {1 + \dfrac{1}{5}} \right) = 750$

$\Rightarrow {5^x}\left( {\dfrac{6}{5}} \right) = 750$

$\Rightarrow {5^x} = \dfrac{{750 \times 5}}{6} \Rightarrow {5^x} = 125 \times 5$

$\Rightarrow {5^x} = {5^3} \times 5$

$\Rightarrow {5^x} = {5^4} \Rightarrow x = 4$

Ans: ${a^b} + {b^a} = {\left( { - 1} \right)^2} + {\left( 2 \right)^{ - 1}}$

= $1 + \dfrac{1}{2} = \dfrac{{2 + 1}}{2} = \dfrac{3}{2}$

(2) ab – ba 

Ans: ${a^b} - {b^a} = {\left( { - 1} \right)^2} - {\left( 2 \right)^{ - 1}}$

= $1 - \dfrac{1}{2} = \dfrac{{2 - 1}}{2} = \dfrac{1}{2}$

(3) ab × b2 

Ans: ${a^b} \times {b^a} = {\left( { - 1} \right)^2} \times {\left( 2 \right)^{ - 1}}$

= $1 \times \dfrac{1}{2} = \dfrac{1}{2}$ 

(4) ab ÷ ba

Ans: ${a^b} \div {b^a} = {\left( { - 1} \right)^2} \div {\left( 2 \right)^{ - 1}}$

= $1 \div \dfrac{1}{2} = 1 \times 2 = 2$

Ans: $\dfrac{{ - 1296}}{{14641}} =  - \dfrac{{{{\left( 6 \right)}^4}}}{{{{\left( {11} \right)}^4}}} = - {\left( {\dfrac{6}{{11}}} \right)^4}$

(2) $\dfrac{{ - 125}}{{343}}$

Ans: $\dfrac{{ - 125}}{{343}} =  - \dfrac{{{{\left( 5 \right)}^3}}}{{{{\left( 7 \right)}^3}}} =  - {\left( {\dfrac{5}{7}} \right)^3}$

(3) $\dfrac{{400}}{{3969}}$

Ans: $\dfrac{{400}}{{3969}} = \dfrac{{{{\left( {20} \right)}^2}}}{{{{\left( {63} \right)}^2}}} = {\left( {\dfrac{{20}}{{63}}} \right)^2}$

(4) $\dfrac{{ - 625}}{{10000}}$

Ans: $\dfrac{{ - 625}}{{10000}} = \dfrac{{ - {{\left( 5 \right)}^4}}}{{{{\left( {10} \right)}^4}}} =  - {\left( {\dfrac{1}{2}} \right)^4}$

Ans: This equation can be simplified as:

${\left( {{{\left( {\dfrac{1}{2}} \right)}^2} - {{\left( {\dfrac{1}{4}} \right)}^3}} \right)^{ - 1}} \times {2^{ - 3}} = {\left( {\dfrac{1}{4} - \dfrac{1}{{64}}} \right)^1} \times {\left( {\dfrac{1}{2}} \right)^3}$

= ${\left( {\dfrac{{15}}{{64}}} \right)^{ - 1}} \times \left( {\dfrac{1}{8}} \right) = \dfrac{{64}}{{15}} \times \dfrac{1}{8} = \dfrac{8}{{15}}$

(2) ${\left( {{{\left( {\dfrac{4}{3}} \right)}^{ - 2}} - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)^{ - 2}}$

Ans: This equation can be simplified as:

${\left( {{{\left( {\dfrac{4}{3}} \right)}^{ - 2}} - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)^{ - 2}} = {\left( {{{\left( {\dfrac{3}{4}} \right)}^2} - {{\left( {\dfrac{3}{4}} \right)}^2}} \right)^{ - 2}} = {\left( 0 \right)^{ - 2}} = 0$

(3) ${\left( {\dfrac{4}{{13}}} \right)^4} \times {\left( {\dfrac{{13}}{7}} \right)^2} \times {\left( {\dfrac{7}{4}} \right)^3}$

Ans: This equation can be simplified as:

${\left( {\dfrac{4}{{13}}} \right)^4} \times {\left( {\dfrac{{13}}{7}} \right)^2} \times {\left( {\dfrac{7}{4}} \right)^3} = {4^4} \times {13^{ - 4}} \times {13^2} \times {7^{ - 2}} \times {7^3} \times {4^{ - 3}}$

=${4^{4 - 3}} \times {13^{2 - 4}} \times {7^{3 - 2}}$

= ${4^1} \times {13^{ - 2}} \times {7^1}$

= $\dfrac{{4 \times 7}}{{{{13}^2}}} = \dfrac{{28}}{{169}}$

(4) ${\left( {\dfrac{1}{5}} \right)^{45}} \times {\left( {\dfrac{1}{5}} \right)^{ - 60}} - {\left( {\dfrac{1}{5}} \right)^{ - 28}} \times {\left( {\dfrac{1}{5}} \right)^{ - 43}}$

Ans: This equation can be simplified as:

${\left( {\dfrac{1}{5}} \right)^{45}} \times {\left( {\dfrac{1}{5}} \right)^{ - 60}} - {\left( {\dfrac{1}{5}} \right)^{28}} \times {\left( {\dfrac{1}{5}} \right)^{ - 43}}$

= ${5^{ - 45}} \times {5^{60}} - {5^{ - 28}} \times {5^{43}}$

= ${5^{ - 45 + 60}} - {5^{ - 28 + 43}} = {5^{15}} - {5^{15}} = 0$

(5) $\dfrac{{{{\left( 9 \right)}^3} \times 27 \times {t^4}}}{{{{\left( 3 \right)}^{ - 2}} \times {{\left( 3 \right)}^4} \times {t^2}}}$

Ans: This equation can be simplified as:

$\dfrac{{{{\left( {{3^2}} \right)}^3} \times {{\left( 3 \right)}^3} \times {t^4}}}{{{{\left( 3 \right)}^{ - 2}} \times {{\left( 3 \right)}^4} \times {t^2}}} = {3^6} \times {3^3} \times {3^2} \times {3^{ - 4}} \times {t^4} \times {t^{ - 2}}$

= ${\left( 3 \right)^{11 - 4}} \times {t^{4 - 2}} = {3^7} \times {t^2}$

(6) $\dfrac{{{{\left( {{3^{ - 2}}} \right)}^2} \times {{\left( {{5^2}} \right)}^{ - 3}} \times {{\left( {{t^{ - 3}}} \right)}^2}}}{{{{\left( {{3^{ - 2}}} \right)}^5} \times {{\left( {{5^3}} \right)}^{ - 2}} \times {{\left( {{t^{ - 4}}} \right)}^3}}}$

Ans: This equation can be simplified as:

$\dfrac{{{{\left( {{3^{ - 2}}} \right)}^2} \times {{\left( {{5^2}} \right)}^{ - 3}} \times {{\left( {{t^{ - 3}}} \right)}^2}}}{{{{\left( {{3^{ - 2}}} \right)}^5} \times {{\left( {{5^3}} \right)}^{ - 2}} \times {{\left( {{t^{ - 4}}} \right)}^3}}} = \dfrac{{{3^{ - 4}} \times {5^{ - 6}} \times {t^{ - 6}}}}{{{3^{ - 10}} \times {5^{ - 6}} \times {t^{ - 12}}}}$

= ${3^{ - 4 + 10}} \times {t^{ - 6 + 12}}$

= ${3^6} \times {t^6}$

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The NCERT Exemplar Maths workbook is one of the workbooks in the NCERT Exemplar series. The NCERT Exemplar series contains only practice questions based on the subject curriculum. These questions are at a higher level than the regular classroom practice questions, which is deliberately done to encourage students to push themselves.