Exponents

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.

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

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.

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

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:

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

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

(iii) We have:

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

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

(iv) We have:

$\phantom{=} \Big(\dfrac{-3}{4}\Big)^3 \times \Big(\dfrac{-5}{2}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^5 \\[1em] = \dfrac{(-3)^3}{4^3} \times \dfrac{(-5)^3}{2^3} \times \dfrac{2^5}{3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^5}{4^3 \times 2^3 \times 3^5} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^{5-3}}{4^3 \times 3^5} \quad \text{[Applying exponent law]} \\[1em] = \dfrac{(-3)^3 \times (-5)^3 \times 2^2}{4^3 \times 3^5} \\[1em] = \dfrac{(-3 \times -3 \times -3) \times (-5 \times -5 \times -5) \times (2 \times 2)}{(4 \times 4 \times 4) \times (3 \times 3 \times 3 \times 3 \times 3)} \\[1em] = \dfrac{(-27) \times -125 \times 4}{64 \times 243} \\[1em] = \dfrac{-1 \times -125 \times 1}{16 \times 9} \quad \text{[Dividing 27 and 243 by 27, 4 and 64 by 4]} \\[1em] = \dfrac{125}{144} \\[1em]$

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:

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

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

(ii) We have:

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

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.

3² + 2³ is equal to :

1. 15
2. 16
3. 17
4. 18

Answer

3² = 3 x 3 = 9

2³ = 2 x 2 x 2 = 8.

3² + 2³ = 9 + 8 = 17.

Hence, option 3 is the correct option.

(7 - 5)⁵ is equal to :

1. 6
2. 8
3. 16
4. 32

Answer

We have:

(7 - 5)⁵

⇒ (7 - 5) = 2

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

Hence, option 4 is the correct option.

(-5)⁴ is equal to :

1. 20
2. -125
3. 625
4. -1024

Answer

Concept:

(-5)⁴ = (-5) x (-5) x (-5) x (-5) = 625.

Hence, option 3 is the correct option.

$\Big(\dfrac{-3}{5}\Big)^5$ is equal to :

1. $\dfrac{-243}{3125}$

2. $\dfrac{-81}{625}$

3. $\dfrac{81}{625}$

4. $\dfrac{512}{1875}$

Answer

If the base is negative and the exponent is odd, the result is negative.

$\Big(\dfrac{-3}{5}\Big)^5 \\[1em] = \dfrac{(-3)^5}{5^5} \\[1em] = \dfrac{(-3) \times (-3) \times (-3) \times (-3) \times (-3)}{5 \times 5 \times 5 \times 5 \times 5} \\[1em] = \dfrac{-243}{3125}$

Hence, option 1 is the correct option.

The value of x such that $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$ is

1. -4
2. -2
3. 0
4. 1

Answer

Given:

$\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8} = \Big(\dfrac{3}{7}\Big)^{2x + 3}$

LHS = $\Big(\dfrac{3}{7}\Big)^3 \times \Big(\dfrac{3}{7}\Big)^{-8}$

$= \dfrac{3^3}{7^3} \times \dfrac{3^{-8}}{7^{-8}} \\[1em] = \dfrac{3^3 \times 3^{-8}}{7^3 \times 7^{-8}} \\[1em] = \dfrac{3^{3+(-8)}}{7^{3+(-8)}} = \dfrac{3^{3-8}}{7^{3-8}} \\[1em] = \dfrac{3^{-5}}{7^{-5}} \\[1em] = \Big(\dfrac{3}{7}\Big)^{-5} \\[1em]$

Now, LHS = $\Big(\dfrac{3}{7}\Big)^{-5}$

As the base of both LHS and RHS is same, let us compare the exponents:

-5 = 2x + 3
2x = -5 - 3
2x = -(5 + 3)
2x = -8
x = $\dfrac{-8}{2}$
x = -4

Hence, option 1 is the correct option.

The value of $\Big[\Big\lbrace\Big(\dfrac{-1}{3}\Big)^{-2}\Big\rbrace^{-1}\Big]^{2}$ is

1. 81

2. -81

3. $\dfrac{1}{81}$

4. $-\dfrac{1}{81}$

Answer

Given:

$\Big[\Big\lbrace\Big(\dfrac{-1}{3}\Big)^{-2}\Big\rbrace^{-1}\Big]^{2}$

Multiply the exponents: (-2) x (-1) x 2 = 4. $\quad$ [Power of power rule]

∴ $\Big(\dfrac{-1}{3}\Big)^4 = \dfrac{(-1)^4}{3^4} = \dfrac{1}{81}$

Hence, option 3 is the correct option.

The number 34613000 when expressed in exponential form is equal to

1. 3461.3 x 10³
2. 3.4613 x 10⁷
3. 0.34613 x 10⁹
4. 34.613 x 10⁵

Answer

Move the decimal 7 places to the left: 3.4613 x 10⁷.

Verification: 3.4613 x 10,000,000 = 34613000.

Hence, option 2 is the correct option.

Which is larger ?

(i) 2³ or 3²

(ii) 2⁵ or 5²

(iii) 3⁷ or 7³.

Answer

(i) 2³ or 3²

2³ = 2 x 2 x 2 = 8

3² = 3 x 3 = 9

Clearly, 8 < 9.

Hence, 3² is larger.

(ii) 2⁵ or 5²

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

5² = 5 x 5 = 25

Clearly, 32 > 25.

Hence, 2⁵ is larger.

(iii) 3⁷ or 7³.

3⁷ = 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187

7³ = 7 x 7 x 7 = 343

Clearly, 2187 > 343.

Hence, 3⁷ is larger.

Fill in the blanks :

(i) In an exponential form x^m; x is called the ............... .

(ii) We have : $\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}$ if ............... .

(iii) Any number to the power 0 is equal to ............... .

(iv) The exponential form is also called ............... .

(v) The reciprocal of x^a is equal to x to the power ............... .

Answer

(i) In an exponential form x^m; x is called the base.

(ii) We have : $\dfrac{a^m}{a^n} = \dfrac{1}{a^{n-m}}$ if n > m.

(iii) Any number to the power 0 is equal to 1.

(iv) The exponential form is also called power notation.

(v) The reciprocal of x^a is equal to x to the power -a.

State True or False :

(i) For any non-zero number a, we have (a^m)ⁿ = a^m+n.

(ii) If a and b are non-zero numbers, then $\Big\lbrace\Big(\dfrac{a}{b}\Big)^m\Big\rbrace^n = \Big(\dfrac{b}{a}\Big)^{-mn}$

(iii) For a non-zero number x; (x^m x xⁿ) is equal to x to the power (m + n).

(iv) If a number p is multiplied n times, then the resulting number is pⁿ.

(v) In an exponential notation xⁿ; n is called the index.

Answer

(i) False
Reason — The power of a power law states that (a^m)ⁿ = a^{m x n}. The expression a^m+n is the result of multiplying powers with the same base (a^m x aⁿ).

(ii) True
Reason — According to the power of a power rule, $\Big\lbrace\left(\dfrac{a}{b}\right)^m\Big\rbrace^n = \left(\dfrac{a}{b}\right)^{mn}$. By applying the reciprocal rule $\left(\dfrac{a}{b}\right)^{mn} = \left(\dfrac{b}{a}\right)^{-mn}$, the statement is mathematically correct.

(iii) True
Reason — This is the product law of exponents, which states that for any non-zero base x, x^m x xⁿ = x^(m+n).

(iv) True
Reason — By definition, exponential notation is a shorthand for repeated multiplication; if p is multiplied n times, it is written as pⁿ.

(v) True
Reason — In the notation xⁿ, the number n is commonly referred to as the exponent, power, or index.

A computer purchased for ₹72900 loses two-third of its value every year. Its value is evaluated at the end of every year.

(1) Which of the following expressions gives the value of the computer (in ₹) after n years?

1. $\dfrac{72900}{\Big(\dfrac{2}{3}\Big)^{n}}$

2. $\dfrac{2^{n} \times 27900}{3^{n}}$

3. $\dfrac{72900}{3^{n}}$

4. $\dfrac{72900}{2^{n} \times 3^{n}}$

(2) Find the value of the computer after 3 years.

1. ₹8100
2. ₹2430
3. ₹5600
4. ₹2700

(3) In how many years will the value of the computer be less than ₹200 ?

1. 6 years
2. 7 years
3. 8 years
4. 10 years

(4) By how much will the value of the computer reduce in 4 years ?

1. ₹24300
2. ₹1800
3. ₹8100
4. ₹72000

Answer

(1) Given:

Initial Value = ₹72,900

Loss in value every year = $\dfrac{2}{3}$

Value remaining every year = $1 - \dfrac{2}{3} = \dfrac{1}{3}$ of the previous year's value.

Every year, the value is multiplied by $\dfrac{1}{3}$. After n years, the value is $72900 \times (\dfrac{1}{3})^n$, which is $\dfrac{72900}{3^n}$.

Hence, option 3 is the correct option.

(2) Value of the computer after 3 years = ?

Value of the computer after n years = $\dfrac{72900}{3^n} \quad$ [From previous step]

By replacing the value of 'n' with 3, we get:

₹ $\dfrac{72900}{3^3} = ₹ \dfrac{72900}{27}$

= ₹2700

Hence, option 4 is the correct option.

(3) We know at 3 years, value = ₹2700. $\quad$ [From previous step]

Value remaining every year = $1 - \dfrac{2}{3} = \dfrac{1}{3}$ of the previous year's value. $\quad$ [From step 1]

∴ Value after 4 years = $\dfrac{1}{3} \times 2700 = \dfrac{2700}{3} = ₹ 900$

Value after 5 years = $\dfrac{1}{3} \times 900 = \dfrac{900}{3} = ₹ 300$

Value after 6 years = $\dfrac{1}{3} \times 300 = \dfrac{300}{3} = ₹ 100$

Since 100 < 200, it takes 6 years.

Hence, option 1 is the correct option.

(4) Value of the computer reduced in 4 years = ?

Value after 4 years = ₹ $\dfrac{72900}{3^4} = ₹ \dfrac{72900}{81} = ₹ 900$

Value of the computer reduced in 4 years = Initial value - Value after 4 years

Substituting the values in above, we get:

Value of the computer reduced in 4 years = ₹72900 - ₹ 900 = ₹ 72000

Hence, option 4 is the correct option.

In a bacteria culture under observation in a laboratory, the population of 50 bacteria doubles itself every hour.

(1) Which of the following expressions gives the bacterial population after n hours ?

1. $\dfrac{50}{2^{n}}$

2. 25ⁿ

3. 50 x 2ⁿ

4. $\dfrac{50^{n}}{2}$

(2) The population size of the bacteria after 3 hours will be

1. 200
2. 300
3. 400
4. 500

(3) How many bacteria will be there in the culture after 1 day ?

1. $\dfrac{50}{2^{12}}$

2. 50 x 2²⁴

3. 50 x 2¹²

4. $\dfrac{50}{2^{24}}$

(4) If the culture is observed after every one hour, find the number of hours after which the population size of the bacteria will be larger than 1000.

1. 5
2. 6
3. 8
4. 10

Answer

(1) Given:

Initial Population = 50

Growth Rate = Doubles every hour (x2)

The population starts at 50 and multiplies by 2 for every hour.

General formula after n hours:

Population = 50 × 2ⁿ

Hence, option 3 is the correct option.

(2) Population size after 3 hours = ?

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 3, we get:

Population size after 3 hours = 50 × 2³ = 50 x 8 = 400

Hence, option 3 is the correct option.

(3) Bacteria in the culture after 1 day = ?

We know that 1 day has 24 hours

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 24, we get:

Bacteria in the culture after 1 day = 50 × 2²⁴

Hence, option 2 is the correct option.

(4) After how many hours will the population be larger than 1000?

Let's test hours (n):

n = 4: 50 x 2⁴ = 50 x 16 = 800

n = 5: 50 x 2⁵ = 50 x 32 = 1600

Since 1600 > 1000, it happens at 5 hours.

Hence, option 1 is the correct option.

Assertion: $\dfrac{x^{4}}{y^{3}}$ can also be written as $\dfrac{x+x+x+x}{y+y+y}$.

Reason: In x^m, x is called the base and m is called the exponent.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false because, the expression $\dfrac{x^4}{y^3}$ represents repeated multiplication, not addition. Specifically, $x^4 = x \times x \times x \times x$ and $y^3 = y \times y \times y$. Adding the variables as shown in the assertion is incorrect.

Reason is true as it is the correct mathematical definition for the components of an exponential term.

Hence, option 4 is the correct option.

Assertion: x^m ÷ y^m = (x ÷ y)^m

Reason: a^m x b^m = (ab)^m

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Assertion is true because, it is the Power of a Quotient rule. It states that when two different bases are divided and raised to the same power, the power can be applied to the quotient.

Reason is true because, it is the Power of a Product rule. It correctly states that a^m x b^m = (ab)^m.

While both are valid laws of exponents, the rule for multiplication (Reason) does not explain the rule for division (Assertion).

Hence, option 2 is the correct option.

Assertion: 2⁰ + 3⁰ + 4⁰ = (2 + 3 + 4)⁰

Reason: x⁰ = 1.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false.

Let's evaluate both sides using the zero exponent rule:

LHS: 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3.

RHS: (2 + 3 + 4)⁰ = (9)⁰ = 1

Since $3 \neq 1$, the assertion is false.

Reason is true because, the rule x⁰ = 1 (for any non-zero x) is a fundamental law of exponents.

Hence, option 4 is the correct option.