Exponents - Exercise 5(A)

Evaluate :

(i) 7⁴

(ii) (-5)³

(iii) $\Big(\dfrac{3}{4}\Big)^5$

(iv) $\Big(\dfrac{-5}{2}\Big)^2$

Answer

(i) 7⁴

7⁴ = 7 x 7 x 7 x 7 = 2401

Hence, the answer is 2401

(ii) (-5)

(-5)³ = (-5) x (-5) x (-5) = -125

Hence, the answer is -125

(iii) $\Big(\dfrac{3}{4}\Big)^5$

We know that

$\Big(\dfrac{a}{b}\Big)^n = \dfrac{a^n}{b^n} \quad \text{[Power of division rule]}$

$\therefore \Big(\dfrac{3}{4}\Big)^5 = \dfrac{3^5}{4^5} = \dfrac{3 \times 3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4 \times 4} = \dfrac{243}{1024}$

Hence, the answer is $\dfrac{243}{1024}$

(iv) $\Big(\dfrac{-5}{2}\Big)^2$

$\Big(\dfrac{a}{b}\Big)^n = \dfrac{a^n}{b^n} \quad \text{[Power of division rule]}$

$\Big(\dfrac{-5}{2}\Big)^2 = \dfrac{(-5)^2}{2^2} = \dfrac{(-5) \times (-5)}{2 \times 2} = \dfrac{25}{4} = 6\dfrac{1}{4}$

Hence, the answer is $6\dfrac{1}{4}$

Write as a power of 10 :

(i) 10000

(ii) One Crore

(iii) One million

Answer

(i) 10000

10000 = 10 x 10 x 10 x 10 = 10⁴.

Hence, the answer is 10⁴

(ii) One Crore

One Crore is written as 1,00,00,000, which has 7 zeros.

1,00,00,000 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10⁷.

Hence, the answer is 10⁷

(iii) One million

One million is written as 1,000,000, which has 6 zeros.

1,000,000 = 10 x 10 x 10 x 10 x 10 x 10 = 10⁶.

Hence, the answer is 10⁶

Express each of the following in exponential notation :

(i) $\Big(\dfrac{-7}{13}\Big) \times \Big(\dfrac{-7}{13}\Big) \times \Big(\dfrac{-7}{13}\Big)$

(ii) $\Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big)$

Answer

(i) $\Big(\dfrac{-7}{13}\Big) \times \Big(\dfrac{-7}{13}\Big) \times \Big(\dfrac{-7}{13}\Big)$

The rational number $\Big(\dfrac{-7}{13}\Big)$ is being multiplied by itself 3 times.

Hence, the answer is $\Big(\dfrac{-7}{13}\Big)^3$

(ii) $\Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big) \times \Big(\dfrac{-8}{3}\Big)$

The rational number $\Big(\dfrac{-8}{3}\Big)$ is being multiplied by itself 4 times.

Hence, the answer is $\Big(\dfrac{-8}{3}\Big)^4$

Express each of the following in exponential notation :

(i) $\dfrac{343}{512}$

(ii) $\dfrac{-32}{243}$

(iii) $\dfrac{-1}{128}$

(iv) $\dfrac{729}{64}$

Answer

(i) $\dfrac{343}{512}$

We have:

343 = 7 x 7 x 7 = 7³

512 = 8 x 8 x 8 = 8³

∴ $\dfrac{343}{512} = \dfrac{7^3}{8^3} = \Big(\dfrac{7}{8}\Big)^3$

Hence, the answer is $\Big(\dfrac{7}{8}\Big)^3$

(ii) $\dfrac{-32}{243}$

We have:

-32 = (-2) x (-2) x (-2) x (-2) x (-2) = (-2)⁵

243 = 3 x 3 x 3 x 3 x 3 = 3⁵

∴ $\dfrac{-32}{243} = \dfrac{(-2)^5}{3^5} = \Big(\dfrac{-2}{3}\Big)^5$

Hence, the answer is $\Big(\dfrac{-2}{3}\Big)^5$

(iii) $\dfrac{-1}{128}$

We have denominator 128:

∴ 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

-1 = (-1)⁷ $\quad$ (Since any odd power of -1 is -1)

∴ $\dfrac{-1}{128} = \dfrac{(-1)^7}{2^7} = \Big(\dfrac{-1}{2}\Big)^7$

Hence, the answer is $\Big(\dfrac{-1}{2}\Big)^7$

(iv) $\dfrac{729}{64}$

We have:

729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

∴ $\dfrac{729}{64} = \Big(\dfrac{3}{2}\Big)^6$

Hence, the answer is $\Big(\dfrac{3}{2}\Big)^6$

Express each of the following in exponential notation :

(i) $\Big(\dfrac{5}{21}\Big)^3 \times \Big(\dfrac{5}{21}\Big)^8$

(ii) $\Big(\dfrac{-7}{3}\Big)^{11} \times \Big(\dfrac{-7}{3}\Big)^{13}$

(iii) $\Big(\dfrac{13}{43}\Big)^7 ÷ \Big(\dfrac{13}{43}\Big)^2$

(iv) $\Big(\dfrac{-16}{35}\Big)^{16} ÷ \Big(\dfrac{-16}{35}\Big)^3$

(v) $\Big(\dfrac{-7}{15}\Big)^{12} ÷ \Big(\dfrac{-7}{15}\Big)^{15}$

(vi) $\Big(\dfrac{1}{24}\Big)^{13} ÷ \Big(\dfrac{1}{24}\Big)^{16}$

Answer

(i) $\Big(\dfrac{5}{21}\Big)^3 \times \Big(\dfrac{5}{21}\Big)^8$

Using the multiplication rule, we add the exponents: 3 + 8 = 11.

$\therefore \Big(\dfrac{5}{21}\Big)^3 \times \Big(\dfrac{5}{21}\Big)^8 \\[1em] = \Big(\dfrac{5}{21}\Big)^{3+8} \\[1em] = \Big(\dfrac{5}{21}\Big)^{11}$

Hence, the answer is $\Big(\dfrac{5}{21}\Big)^{11}$

(ii) $\Big(\dfrac{-7}{3}\Big)^{11} \times \Big(\dfrac{-7}{3}\Big)^{13}$

Using the multiplication rule, we add the exponents: 11 + 13 = 24.

$\therefore \Big(\dfrac{-7}{3}\Big)^{11} \times \Big(\dfrac{-7}{3}\Big)^{13} \\[1em] = \Big(\dfrac{-7}{3}\Big)^{11+13} \\[1em] = \Big(\dfrac{-7}{3}\Big)^{24}$

Hence, the answer is $\Big(\dfrac{-7}{3}\Big)^{24}$

(iii) $\Big(\dfrac{13}{43}\Big)^7 ÷ \Big(\dfrac{13}{43}\Big)^2$

Using the division rule, we subtract the exponents: 7 - 2 = 5.

$\therefore \Big(\dfrac{13}{43}\Big)^7 ÷ \Big(\dfrac{13}{43}\Big)^2 \\[1em] = \Big(\dfrac{13}{43}\Big)^{7-2} \\[1em] = \Big(\dfrac{13}{43}\Big)^5$

Hence, the answer is $\Big(\dfrac{13}{43}\Big)^5$

(iv) $\Big(\dfrac{-16}{35}\Big)^{16} ÷ \Big(\dfrac{-16}{35}\Big)^3$

Using the division rule, we subtract the exponents: 16 - 3 = 13.

$\therefore \Big(\dfrac{-16}{35}\Big)^{16} ÷ \Big(\dfrac{-16}{35}\Big)^3 \\[1em] = \Big(\dfrac{-16}{35}\Big)^{16-3} \\[1em] = \Big(\dfrac{-16}{35}\Big)^{13}$

Hence, the answer is $\Big(\dfrac{-16}{35}\Big)^{13}$

(v) $\Big(\dfrac{-7}{15}\Big)^{12} ÷ \Big(\dfrac{-7}{15}\Big)^{15}$

Using the division rule, we subtract the exponents: 12 - 15 = -3.

$\therefore \Big(\dfrac{-7}{15}\Big)^{12} ÷ \Big(\dfrac{-7}{15}\Big)^{15} \\[1em] = \Big(\dfrac{-7}{15}\Big)^{12-15} \\[1em] = \Big(\dfrac{-7}{15}\Big)^{-3} \quad \text{[Power is negative, take reciprocal]} \\[1em] = \Big(\dfrac{-15}{7}\Big)^3$

Hence, the answer is $\Big(\dfrac{-15}{7}\Big)^3$

(vi) $\Big(\dfrac{1}{24}\Big)^{13} ÷ \Big(\dfrac{1}{24}\Big)^{16}$

Using the division rule, we subtract the exponents: 13 - 16 = -3.

$\therefore \Big(\dfrac{1}{24}\Big)^{13} ÷ \Big(\dfrac{1}{24}\Big)^{16} \\[1em] = \Big(\dfrac{1}{24}\Big)^{13-16} \\[1em] = \Big(\dfrac{1}{24}\Big)^{-3} \quad \text{[Power is negative, take the reciprocal]} \\[1em] = 24^3$

Hence, the answer is 24³

Simplify and express each of the following as a rational number :

(i) $\Big(\dfrac{6}{5}\Big)^3 \times \Big(\dfrac{5}{2}\Big)^2$

(ii) $\Big(\dfrac{3}{4}\Big)^2 \times \Big(\dfrac{-1}{2}\Big)^5 \times 2^3$

(iii) $\Big(\dfrac{5}{4}\Big)^2 \times \Big(\dfrac{2}{3}\Big)^2 \times \Big(\dfrac{-3}{5}\Big)^3$

(iv) $\Big(\dfrac{-3}{4}\Big)^3 \times \Big(\dfrac{-5}{2}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^5$

(v) $\Big(\dfrac{7}{11}\Big)^6 ÷ \Big(\dfrac{7}{11}\Big)^3$

(vi) $\Big(\dfrac{-4}{3}\Big)^8 ÷ \Big(\dfrac{-4}{3}\Big)^{12}$

Answer

(i) We have:

$\phantom{=} \Big(\dfrac{6}{5}\Big)^3 \times \Big(\dfrac{5}{2}\Big)^2 \\[1em] = \dfrac{6^3}{5^3} \times \dfrac{5^2}{2^2} \\[1em] = \dfrac{6^3 \times 5^2}{5^3 \times 2^2} \\[1em] = \dfrac{6^3 \times 5^{2-3}}{2^2}\quad \text{[Applying exponent law]} \\[1em] = \dfrac{6^3 \times 5^{-1}}{2^2} \\[1em] = \dfrac{6 \times 6 \times 6}{5 \times 2 \times 2} \quad {[5^{-1} = \dfrac{1}{5}]} \\[1em] = \dfrac{216}{20} \\[1em] = \dfrac{54}{5} = 10\dfrac{4}{5}$

Hence, the answer is $10\dfrac{4}{5}$

(ii) We have:

$\phantom{=} \Big(\dfrac{3}{4}\Big)^2 \times \Big(\dfrac{-1}{2}\Big)^5 \times 2^3 \\[1em] = \dfrac{3^2}{4^2} \times \dfrac{(-1)^5}{2^5} \times 2^3 \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^3}{4^2 \times 2^5} \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^{3-5}}{4^2} \\[1em] = \dfrac{3^2 \times (-1)^5 \times 2^{-2}}{4^2} \\[1em] = \dfrac{(3 \times 3) \times (-1)}{4 \times (4 \times 4)} \quad [2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}] \\[1em] = \dfrac{-9}{64}$

Hence, the answer is $\dfrac{-9}{64}$

(iii) We have:

$\phantom{=} \Big(\dfrac{5}{4}\Big)^2 \times \Big(\dfrac{2}{3}\Big)^2 \times \Big(\dfrac{-3}{5}\Big)^3 \\[1em] \dfrac{5^2}{4^2} \times \dfrac{2^2}{3^2} \times \dfrac{(-3)^3}{5^3} \\[1em] = \dfrac{5^2 \times 2^2 \times (-3)^3}{4^2 \times 3^2 \times 5^3} \\[1em] = \dfrac{2^2 \times (-3)^{3-2} \times 5^{2-3}}{4^2} \\[1em] = \dfrac{2^2 \times (-3)^1 \times 5^{-1}}{4^2} \\[1em] = \dfrac{(2 \times 2) \times (-3)}{(4 \times 4) \times 5} \quad[5^{-1} = \dfrac{1}{5}] \\[1em] = \dfrac{-12}{80} = \dfrac{-3}{20}$

Hence, the answer is $\dfrac{-3}{20}$

(iv) We have:

$\phantom{=} \Big(\dfrac{-3}{4}\Big)^3 \times \Big(\dfrac{-5}{2}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^5 \\[1em] = \dfrac{(-3)^3}{4^3} \times \dfrac{(-5)^3}{2^3} \times \dfrac{2^5}{3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^5}{4^3 \times 2^3 \times 3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^{5-3}}{4^3 \times 3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^2}{4^3 \times 3^5} \\[1em] = \dfrac{(-3 \times -3 \times -3) \times (-5 \times -5 \times -5) \times (2 \times 2)}{(4 \times 4 \times 4) \times (3 \times 3 \times 3 \times 3 \times 3)} \\[1em] = \dfrac{(-27) \times -125 \times 4}{64 \times 243} \\[1em] = \dfrac{-1 \times -125 \times 1}{16 \times 9} \\[1em] = \dfrac{125}{144} \\[1em]$

Hence, the answer is $\dfrac{125}{144}$

(v) We have:

$\phantom{=} \Big(\dfrac{7}{11}\Big)^6 ÷ \Big(\dfrac{7}{11}\Big)^3 \\[1em] = \Big(\dfrac{7}{11}\Big)^{6-3} \quad \text{[By division rule]} \\[1em] = \Big(\dfrac{7}{11}\Big)^3 \\[1em] = \dfrac{7^3}{11^3} \\[1em] = \dfrac{7 \times 7 \times 7}{11 \times 11 \times 11} \\[1em] = \dfrac{343}{1331}$

Hence, the answer is $\dfrac{343}{1331}$

(vi) We have:

$\phantom{=} \Big(\dfrac{-4}{3}\Big)^8 ÷ \Big(\dfrac{-4}{3}\Big)^{12} \\[1em] = \Big(\dfrac{-4}{3}\Big)^{8-12} \quad \text{[By division rule]} \\[1em] = \Big(\dfrac{-4}{3}\Big)^{-4} \\[1em] = \Big(\dfrac{3}{-4}\Big)^{4} \quad \text{[By reciprocal rule]} \\[1em] = \dfrac{3 \times 3 \times 3 \times 3}{(-4) \times (-4) \times (-4) \times (-4)} \\[1em] = \dfrac{81}{256}$

Hence, the answer is $\dfrac{81}{256}$

Simplify and express each of the following as a rational number :

(i) $\dfrac{10^2 \times 15^3}{2^2 \times 3 \times 5^5 \times 6^4}$

(ii) $\dfrac{3^5 \times 25 \times 10^5}{5^7 \times 6^5}$

Answer

(i) We have:

$\phantom{=} \dfrac{10^2 \times 15^3}{2^2 \times 3 \times 5^5 \times 6^4} \\[1em] = \dfrac{(2 \times 5)^2 \times (3 \times 5)^3}{2^2 \times 3^1 \times 5^5 \times (2 \times 3)^4} \\[1em] = \dfrac{2^2 \times 5^2 \times 3^3 \times 5^3}{2^2 \times 3^1 \times 5^5 \times 2^4 \times 3^4} \\[1em] = \dfrac{2^2 \times 3^3 \times 5^{2+3}}{2^{2+4} \times 3^{1+4} \times 5^5} \\[1em] = \dfrac{2^2 \times 3^3 \times 5^5}{2^6 \times 3^5 \times 5^5} \\[1em] = \dfrac{2^{2-6}}{3^{5-3} \times 5^{5-5}} \\[1em] = \dfrac{2^{-4}}{3^2 \times 5^0} \quad \text{[Using law of exponents]} \\[1em] = \dfrac{1}{2^4 \times 3^2 \times 1} \quad [2^{-4} = \dfrac{1}{2^4}] \\[1em] = \dfrac{1}{16 \times 9} \\[1em] = \dfrac{1}{144}$

Hence, the answer is $\dfrac{1}{144}$

(ii) We have:

$\phantom{=} \dfrac{3^5 \times 25 \times 10^5}{5^7 \times 6^5} \\[1em] = \dfrac{3^5 \times 5^2 \times (2 \times 5)^5}{5^7 \times (2 \times 3)^5} \\[1em] = \dfrac{3^5 \times 5^2 \times 2^5 \times 5^5}{5^7 \times 2^5 \times 3^5} \\[1em] = \dfrac{2^5 \times 3^5 \times 5^{2+5}}{2^5 \times 3^5 \times 5^7} \\[1em] = \dfrac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7} \\[1em] = 2^{5-5} \times 3^{5-5} \times 5^{7-7} \quad \text{[law of exponents]} \\[1em] = 2^0 \times 3^0 \times 5^0 \\[1em] = 1 \times 1 \times 1 \\[1em] = 1$

Hence, the answer is 1

The distance between the Earth and the Moon is approximately 384000 km. Express this distance in metres in exponential notation.

Answer

Given:

Distance between the Earth and the Moon = 384000 km.

We know that 1 km = 1000 m = 10³ m.

Distance in metres = 384000 x 10³ m.

We can write 384000 as 384 x 1000 = 384 x 10³.

Total distance = (384 x 10³) x 10³ m.

Using the law of exponents (a^m x aⁿ = a^m+n), we add the powers: 3+3 = 6.

Distance = 384 x 10⁶ m.

Hence, the distance is 384 x 10⁶ m.

The RAM of a computer is 8 gigabyte. If each gigabyte is equal to 10⁹ bytes, then express the RAM in bytes.

Answer

Given:

Total RAM = 8 gigabytes.

Value of 1 gigabyte = 10⁹ bytes.

RAM in bytes = Total RAM x Value of 1 gigabyte

Substituting the values in above, we get:

RAM in bytes = 8 x 10⁹ bytes.

Hence, the RAM is 8 x 10⁹ bytes.

In a tennis competition, 128 players were selected for a series of knockout rounds. In each round the losers were eliminated and the winners reached the next round. How many players moved to the next round after 4th round? Express this number in the exponential notation in terms of the initial number of players.

Answer

Given:

Initial number of players = 128.

In a knockout round, the number of players is reduced to half $\Big(\dfrac{1}{2}\Big)$.

After 1st round, players left = $\dfrac{1}{2} \times 128$.

After 2nd round, players left = $\dfrac{1}{2} \times \dfrac{1}{2} \times 128 = \Big(\dfrac{1}{2}\Big)^2 \times 128$.

After 3rd round, players left = $\Big(\dfrac{1}{2}\Big)^3 \times 128$.

After 4th round, players left = $\Big(\dfrac{1}{2}\Big)^4 \times 128$ $= \dfrac{1^4}{2^4} \times 128 \\[1em] = \dfrac{128}{2^4} \\[1em] = \dfrac{128}{2 \times 2 \times 2 \times 2} \\[1em] = \dfrac{128}{16} \\[1em] = 8$

Hence, 8 players moved to the next round.

Exponential notation in initial no. of players = $\dfrac{\bold{128}}{\bold{2}^\bold{4}}$

Express the following in centimetres (cm) in exponential notation :

(i) 98 hm

(ii) 156 km

(iii) 371 m

Answer

(i) 98 hm

1 hm = 100 m and 1 m = 100 cm.

Therefore, 1 hm = 100 x 100 cm = 10000 cm.

In exponential form, 10000 = 10⁴ cm.

98 hm = 98 x 10⁴ cm.

Hence, the answer is 98 x 10⁴ cm.

(ii) 156 km

1 km = 1000 m and 1 m = 100 cm.

Therefore, 1 km = 1000 x 100 cm = 100000 cm.

In exponential form, 100000 = 10⁵ cm.

156 km = 156 x 10⁵ cm.

Hence, the answer is 156 x 10⁵ cm.

(iii) 371 m

1 m = 100 cm.

In exponential form, 100 = 10² cm.

371 m = 371 x 10² cm.

Hence, the answer is 371 x 10² cm.