Exponents and Powers

Fill in the blanks:

(i) In the expression 3⁷, base = ............... and exponent = ...............

(ii) In the expression (-7)⁵, base = ............... and exponent = ...............

(iii) In the expression $\Big(\dfrac{2}{5}\Big)^{11}$, base = ............... and exponent = ...............

(iv) If base is 6 and exponent is 8, then exponential form = ...............

Answer

As we know, in the expression aⁿ, a is called the base and n is called the exponent (or index).

(i) In 3⁷, base = 3 and exponent = 7

(ii) In (-7)⁵, base = -7 and exponent = 5

(iii) In $\Big(\dfrac{2}{5}\Big)^{11}$, base = $\mathbf{\dfrac{2}{5}}$ and exponent = 11

(iv) If base is 6 and exponent is 8, then exponential form = 6⁸

Find the value of the following:

(i) 2⁶

(ii) 5⁵

(iii) (-6)⁴

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

(v) $\Big(-\dfrac{2}{3}\Big)^5$

(vi) (-2)⁹

Answer

(i) Solving,

⇒ 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64.

Hence, 2⁶ = 64.

(ii) Solving,

⇒ 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125.

Hence, 5⁵ = 3125.

(iii) Solving,

⇒ (-6)⁴ = (-6) × (-6) × (-6) × (-6) = 1296.

Hence, (-6)⁴ = 1296.

(iv) Solving,

$\Big(\dfrac{2}{3}\Big)^4 = \dfrac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} = \dfrac{16}{81}$

Hence, $\Big(\dfrac{2}{3}\Big)^4 = \dfrac{16}{81}$.

(v) Solving,

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

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

(vi) Solving,

⇒ (-2)⁹ = (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) × (-2) = -512.

Hence, (-2)⁹ = -512.

Express the following in the exponential form:

(i) 6 × 6 × 6 × 6 × 6

(ii) t × t × t

(iii) 2 × 2 × a × a × a × a

(iv) a × a × a × c × c × c × c × d

Answer

(i) 6 × 6 × 6 × 6 × 6 = 6⁵

(ii) t × t × t = t³

(iii) 2 × 2 × a × a × a × a = 2²a⁴

(iv) a × a × a × c × c × c × c × d = a³c⁴d¹

Simplify the following:

(i) 7 × 10³

(ii) 2⁵ × 9

(iii) 3³ × 10⁴

Answer

(i) Solving,

⇒ 7 × 10³ = 7 × 10 × 10 × 10 = 7000.

Hence, 7 × 10³ = 7000.

(ii) Solving,

⇒ 2⁵ × 9 = 2 × 2 × 2 × 2 × 2 × 9 = 32 × 9 = 288.

Hence, 2⁵ × 9 = 288.

(iii) Solving,

⇒ 3³ × 10⁴ = (3 × 3 × 3) × (10 × 10 × 10 × 10) = 27 × 10000 = 270000.

Hence, 3³ × 10⁴ = 270000.

Simplify the following:

(i) (-3) × (-2)³

(ii) (-3)² × (-5)²

(iii) (-2)³ × (-10)⁴

(iv) (-1)⁹

(v) 25² × (-1)³¹

(vi) 4² × 3³ × (-1)¹²²

Answer

(i) Solving,

⇒ (-3) × (-2)³

⇒ (-3) × (-8)

⇒ 24.

Hence, (-3) × (-2)³ = 24.

(ii) Solving,

⇒ (-3)² × (-5)²

⇒ [(-3) × (-3)] × [(-5) × (-5)]

⇒ 9 × 25

⇒ 225.

Hence, (-3)² × (-5)² = 225.

(iii) Solving,

⇒ (-2)³ × (-10)⁴

⇒ [(-2) × (-2) × (-2)] × [(-10) × (-10) × (-10) × (-10)]

⇒ (-8) × 10000

⇒ -80000.

Hence, (-2)³ × (-10)⁴ = -80000.

(iv) Solving,

As 9 is an odd natural number,

⇒ (-1)⁹ = -1.

Hence, (-1)⁹ = -1.

(v) Solving,

As 31 is an odd natural number, (-1)³¹ = -1. So,

⇒ 25² × (-1)³¹

⇒ (25 × 25) × (-1)

⇒ 625 × (-1)

⇒ -625.

Hence, 25² × (-1)³¹ = -625.

(vi) Solving,

As 122 is an even natural number, (-1)¹²² = 1. So,

⇒ 4² × 3³ × (-1)¹²²

⇒ (4 × 4) × (3 × 3 × 3) × 1

⇒ 16 × 27 × 1

⇒ 432.

Hence, 4² × 3³ × (-1)¹²² = 432.

Identify the greater number in each of the following:

(i) 4³ or 3⁴

(ii) 7³ or 3⁷

(iii) 4⁵ or 5⁴

(iv) 2¹⁰ or 10²

Answer

(i) Solving,

⇒ 4³ = 4 × 4 × 4 = 64

⇒ 3⁴ = 3 × 3 × 3 × 3 = 81.

Since 81 > 64, therefore 3⁴ > 4³.

Hence, the greater number is 3⁴.

(ii) Solving,

⇒ 7³ = 7 × 7 × 7 = 343

⇒ 3⁷ = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187.

Since 2187 > 343, therefore 3⁷ > 7³.

Hence, the greater number is 3⁷.

(iii) Solving,

⇒ 4⁵ = 4 × 4 × 4 × 4 × 4 = 1024

⇒ 5⁴ = 5 × 5 × 5 × 5 = 625.

Since 1024 > 625, therefore 4⁵ > 5⁴.

Hence, the greater number is 4⁵.

(iv) Solving,

⇒ 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

⇒ 10² = 10 × 10 = 100.

Since 1024 > 100, therefore 2¹⁰ > 10².

Hence, the greater number is 2¹⁰.

Write the following numbers as powers of 2:

(i) 8

(ii) 128

(iii) 1024

Answer

(i) Solving,

⇒ 8 = 2 × 2 × 2 = 2³.

Hence, 8 = 2³.

(ii) Solving,

⇒ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷.

Hence, 128 = 2⁷.

(iii) Solving,

⇒ 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰.

Hence, 1024 = 2¹⁰.

To what power (-2) should be raised to get 16?

Answer

Let the required power be x. Then,

⇒ (-2)^x = 16

⇒ (-2)^x = (-2) × (-2) × (-2) × (-2)

⇒ (-2)^x = (-2)⁴

⇒ x = 4.

Hence, (-2) should be raised to the power 4 to get 16.

Write the following numbers as powers of (-3):

(i) 9

(ii) -27

(iii) 81

Answer

(i) Solving,

⇒ 9 = (-3) × (-3) = (-3)².

Hence, 9 = (-3)².

(ii) Solving,

⇒ -27 = (-3) × (-3) × (-3) = (-3)³.

Hence, -27 = (-3)³.

(iii) Solving,

⇒ 81 = (-3) × (-3) × (-3) × (-3) = (-3)⁴.

Hence, 81 = (-3)⁴.

Find the value of x in each of the following:

(i) 7^x = 343

(ii) 3^x = 729

(iii) (-8)^x = -512

(iv) (-4)^x = -1024

(v) $\Big(\dfrac{2}{5}\Big)^x = \dfrac{32}{3125}$

(vi) $\Big(-\dfrac{3}{4}\Big)^x = -\dfrac{243}{1024}$

Answer

(i) Solving,

⇒ 7^x = 343

⇒ 7^x = 7 × 7 × 7

⇒ 7^x = 7³

⇒ x = 3.

Hence, x = 3.

(ii) Solving,

⇒ 3^x = 729

⇒ 3^x = 3 × 3 × 3 × 3 × 3 × 3

⇒ 3^x = 3⁶

⇒ x = 6.

Hence, x = 6.

(iii) Solving,

⇒ (-8)^x = -512

⇒ (-8)^x = (-8) × (-8) × (-8)

⇒ (-8)^x = (-8)³

⇒ x = 3.

Hence, x = 3.

(iv) Solving,

⇒ (-4)^x = -1024

⇒ (-4)^x = (-4) × (-4) × (-4) × (-4) × (-4)

⇒ (-4)^x = (-4)⁵

⇒ x = 5.

Hence, x = 5.

(v) Solving,

$\Big(\dfrac{2}{5}\Big)^x = \dfrac{32}{3125}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^x = \dfrac{2 \times 2 \times 2 \times 2 \times 2}{5 \times 5 \times 5 \times 5 \times 5}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^x = \Big(\dfrac{2}{5}\Big)^5\\[1em] \Rightarrow x = 5$

Hence, x = 5.

(vi) Solving,

$\Big(-\dfrac{3}{4}\Big)^x = -\dfrac{243}{1024}\\[1em] \Rightarrow \Big(-\dfrac{3}{4}\Big)^x = \dfrac{(-3) \times (-3) \times (-3) \times (-3) \times (-3)}{4 \times 4 \times 4 \times 4 \times 4}\\[1em] \Rightarrow \Big(-\dfrac{3}{4}\Big)^x = \Big(-\dfrac{3}{4}\Big)^5\\[1em] \Rightarrow x = 5$

Hence, x = 5.

Write the prime factorization of the following numbers in the exponential form:

(i) 72

(ii) 360

(iii) 405

(iv) 540

(v) 2280

(vi) 3600

(vii) 4725

(viii) 8400

Answer

(i) By prime factorisation of 72:

$\begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

⇒ 72 = 2 × 2 × 2 × 3 × 3

⇒ 2³ × 3².

Hence, 72 = 2³ × 3².

(ii) By prime factorisation of 360:

$\begin{array}{l|r} 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 360 = 2 × 2 × 2 × 3 × 3 × 5

⇒ 2³ × 3² × 5¹.

Hence, 360 = 2³ × 3² × 5¹.

(iii) By prime factorisation of 405:

$\begin{array}{l|r} 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 405 = 3 × 3 × 3 × 3 × 5

⇒ 3⁴ × 5¹.

Hence, 405 = 3⁴ × 5¹.

(iv) By prime factorisation of 540:

$\begin{array}{l|r} 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 540 = 2 × 2 × 3 × 3 × 3 × 5

⇒ 2² × 3³ × 5¹.

Hence, 540 = 2² × 3³ × 5¹.

(v) By prime factorisation of 2280:

$\begin{array}{l|r} 2 & 2280 \\ \hline 2 & 1140 \\ \hline 2 & 570 \\ \hline 3 & 285 \\ \hline 5 & 95 \\ \hline 19 & 19 \\ \hline & 1 \end{array}$

⇒ 2280 = 2 × 2 × 2 × 3 × 5 × 19

⇒ 2³ × 3¹ × 5¹ × 19¹.

Hence, 2280 = 2³ × 3¹ × 5¹ × 19¹.

(vi) By prime factorisation of 3600:

$\begin{array}{l|r} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

⇒ 2⁴ × 3² × 5².

Hence, 3600 = 2⁴ × 3² × 5².

(vii) By prime factorisation of 4725:

$\begin{array}{l|r} 3 & 4725 \\ \hline 3 & 1575 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

⇒ 4725 = 3 × 3 × 3 × 5 × 5 × 7

⇒ 3³ × 5² × 7¹.

Hence, 4725 = 3³ × 5² × 7¹.

(viii) By prime factorisation of 8400:

$\begin{array}{l|r} 2 & 8400 \\ \hline 2 & 4200 \\ \hline 2 & 2100 \\ \hline 2 & 1050 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

⇒ 8400 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 7

⇒ 2⁴ × 3¹ × 5² × 7¹.

Hence, 8400 = 2⁴ × 3¹ × 5² × 7¹.

Using laws of exponents, simplify and write the following in the exponential form:

(i) 2⁷ × 2⁴

(ii) p⁵ × p³

(iii) (-7)⁵ × (-7)¹¹

(iv) $\Big(\dfrac{3}{5}\Big)^6 \div \Big(\dfrac{3}{5}\Big)^2$

(v) (-6)⁷ ÷ (-6)³

(vi) 7^x × 7³

Answer

As we know, for any rational number a and natural numbers m, n :

a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m-n.

(i) Solving,

⇒ 2⁷ × 2⁴ = 2⁷⁺⁴ = 2¹¹.

Hence, 2⁷ × 2⁴ = 2¹¹.

(ii) Solving,

⇒ p⁵ × p³ = p⁵⁺³ = p⁸.

Hence, p⁵ × p³ = p⁸.

(iii) Solving,

⇒ (-7)⁵ × (-7)¹¹ = (-7)⁵⁺¹¹ = (-7)¹⁶.

Hence, (-7)⁵ × (-7)¹¹ = (-7)¹⁶.

(iv) Solving,

$\Big(\dfrac{3}{5}\Big)^6 ÷ \Big(\dfrac{3}{5}\Big)^2 = \Big(\dfrac{3}{5}\Big)^{6-2} = \Big(\dfrac{3}{5}\Big)^4$

Hence, $\Big(\dfrac{3}{5}\Big)^6 ÷ \Big(\dfrac{3}{5}\Big)^2 = \Big(\dfrac{3}{5}\Big)^4$.

(v) Solving,

⇒ (-6)⁷ ÷ (-6)³ = (-6)^7-3 = (-6)⁴.

Hence, (-6)⁷ ÷ (-6)³ = (-6)⁴.

(vi) Solving,

⇒ 7^x × 7³ = 7^x+3.

Hence, 7^x × 7³ = 7^{x + 3}.

Simplify and write the following in the exponential form:

(i) 5³ × 5⁷ × 5¹²

(ii) a⁵ × a³ × a⁷

(iii) (7¹² × 7³) ÷ 7⁴

Answer

(i) Solving,

⇒ 5³ × 5⁷ × 5¹² = 5³⁺⁷⁺¹² = 5²².

Hence, 5³ × 5⁷ × 5¹² = 5²².

(ii) Solving,

⇒ a⁵ × a³ × a⁷ = a⁵⁺³⁺⁷ = a¹⁵.

Hence, a⁵ × a³ × a⁷ = a¹⁵.

(iii) Solving,

⇒ (7¹² × 7³) ÷ 7⁴

⇒ 7¹²⁺³ ÷ 7⁴

⇒ 7¹⁵ ÷ 7⁴

⇒ 7^15-4

⇒ 7¹¹.

Hence, (7¹² × 7³) ÷ 7⁴ = 7¹¹.

Simplify and write the following in the exponential form:

(i) (2²)¹⁰⁰

(ii) ((-7)⁶)⁵

(iii) (3²)⁵ × (3⁴)⁷

Answer

As we know, for any rational number a and natural numbers m, n : (a^m)ⁿ = a^{m × n}.

(i) Solving,

⇒ (2²)¹⁰⁰ = 2^{2 × 100} = 2²⁰⁰.

Hence, (2²)¹⁰⁰ = 2²⁰⁰.

(ii) Solving,

⇒ ((-7)⁶)⁵ = (-7)^{6 × 5} = (-7)³⁰.

Hence, ((-7)⁶)⁵ = (-7)³⁰.

(iii) Solving,

⇒ (3²)⁵ × (3⁴)⁷

⇒ 3^{2 × 5} × 3^{4 × 7}

⇒ 3¹⁰ × 3²⁸

⇒ 3¹⁰⁺²⁸

⇒ 3³⁸.

Hence, (3²)⁵ × (3⁴)⁷ = 3³⁸.

Simplify and write in the exponential form:

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

(ii) (2³)⁴ ÷ 2⁵

(iii) [(6²)³ ÷ 6³] × 6⁵

Answer

(i) Solving,

$\dfrac{a^3 \times a^5}{(a^3)^2}\\[1em] = \dfrac{a^{3+5}}{a^{3 \times 2}}\\[1em] = \dfrac{a^8}{a^6}\\[1em] = a^{8-6}\\[1em] = a^2$

Hence, $\dfrac{a^3 \times a^5}{(a^3)^2} = a^2$.

(ii) Solving,

⇒ (2³)⁴ ÷ 2⁵

⇒ 2^{3 × 4} ÷ 2⁵

⇒ 2¹² ÷ 2⁵

⇒ 2^12-5

⇒ 2⁷.

Hence, (2³)⁴ ÷ 2⁵ = 2⁷.

(iii) Solving,

⇒ [(6²)³ ÷ 6³] × 6⁵

⇒ [6^{2 × 3} ÷ 6³] × 6⁵

⇒ [6⁶ ÷ 6³] × 6⁵

⇒ 6^6-3 × 6⁵

⇒ 6³ × 6⁵

⇒ 6³⁺⁵

⇒ 6⁸.

Hence, [(6²)³ ÷ 6³] × 6⁵ = 6⁸.

Simplify and write in the exponential form:

(i) 5⁴ × 8⁴

(ii) (-3)⁶ × (-5)⁶

(iii) $\Big(\dfrac{3}{10}\Big)^5 \times \Big(\dfrac{2}{15}\Big)^5$

Answer

As we know, for rational numbers a, b and natural number n : aⁿ × bⁿ = (ab)ⁿ.

(i) Solving,

⇒ 5⁴ × 8⁴ = (5 × 8)⁴ = 40⁴.

Hence, 5⁴ × 8⁴ = 40⁴.

(ii) Solving,

⇒ (-3)⁶ × (-5)⁶ = [(-3) × (-5)]⁶ = 15⁶.

Hence, (-3)⁶ × (-5)⁶ = 15⁶.

(iii)

$\Big(\dfrac{3}{10}\Big)^5 \times \Big(\dfrac{2}{15}\Big)^5\\[1em] = \Big(\dfrac{3}{10} \times \dfrac{2}{15}\Big)^5\\[1em] = \Big(\dfrac{6}{150}\Big)^5\\[1em] = \Big(\dfrac{1}{25}\Big)^5\\[1em] = \Big(\dfrac{1}{5^2}\Big)^5\\[1em] = \Big(\dfrac{1}{5}\Big)^{10}$

Hence, $\Big(\dfrac{3}{10}\Big)^5 \times \Big(\dfrac{2}{15}\Big)^5 = \Big(\dfrac{1}{5}\Big)^{10}$.

Simplify and express each of the following in the exponential form:

(i) $\dfrac{2^4 \times 2 \times 7^3 \times 7^6}{2^3 \times 7^4}$

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

(iii) $\dfrac{2^8 \times a^5}{4^3 \times a^3}$

(iv) $\dfrac{3 \times 7^2 \times 11^8}{21 \times 11^3}$

(v) (2⁰ + 3⁰) 4⁰

(vi) 3⁰ × 4⁰ × 5⁰

Answer

(i) Solving,

$\dfrac{2^4 \times 2 \times 7^3 \times 7^6}{2^3 \times 7^4}\\[1em] = \dfrac{2^{4+1} \times 7^{3+6}}{2^3 \times 7^4}\\[1em] = \dfrac{2^5 \times 7^9}{2^3 \times 7^4}\\[1em] = 2^{5-3} \times 7^{9-4}\\[1em] = 2^2 \times 7^5$

Hence, $\dfrac{2^4 \times 2 \times 7^3 \times 7^6}{2^3 \times 7^4} = 2^2 \times 7^5$.

(ii) Solving,

$\dfrac{(3^2)^3 \times (-2)^5}{(-2)^3}\\[1em] = \dfrac{3^{2 \times 3} \times (-2)^5}{(-2)^3}\\[1em] = 3^6 \times (-2)^{5-3}\\[1em] = 3^6 \times (-2)^2\\[1em] = 3^6 \times 2^2$

Hence, $\dfrac{(3^2)^3 \times (-2)^5}{(-2)^3} = 3^6 \times 2^2$.

(iii) Solving,

$\dfrac{2^8 \times a^5}{4^3 \times a^3}\\[1em] = \dfrac{2^8 \times a^5}{(2^2)^3 \times a^3}\\[1em] = \dfrac{2^8 \times a^5}{2^6 \times a^3}\\[1em] = 2^{8-6} \times a^{5-3}\\[1em] = 2^2 \times a^2\\[1em] = (2a)^2$

Hence, $\dfrac{2^8 \times a^5}{4^3 \times a^3} = (2a)^2$.

(iv) Solving,

$\dfrac{3 \times 7^2 \times 11^8}{21 \times 11^3}\\[1em] = \dfrac{3 \times 7^2 \times 11^8}{3 \times 7 \times 11^3}\\[1em] = 3^{1-1} \times 7^{2-1} \times 11^{8-3}\\[1em] = 3^0 \times 7^1 \times 11^5\\[1em] = 7^1 \times 11^5$

Hence, $\dfrac{3 \times 7^2 \times 11^8}{21 \times 11^3} = 7^1 \times 11^5$.

(v) Solving,

⇒ (2⁰ + 3⁰) 4⁰

⇒ (1 + 1) × 1

⇒ 2 × 1

⇒ 2

⇒ 2¹.

Hence, (2⁰ + 3⁰) 4⁰ = 2¹.

(vi) Solving,

⇒ 3⁰ × 4⁰ × 5⁰

⇒ 1 × 1 × 1

⇒ 1

⇒ 1¹.

Hence, 3⁰ × 4⁰ × 5⁰ = 1¹.

Express each of the following rational numbers in the exponential form:

(i) $\dfrac{25}{64}$

(ii) $-\dfrac{125}{216}$

(iii) $-\dfrac{343}{729}$

Answer

(i) Solving,

$\dfrac{25}{64} = \dfrac{5 \times 5}{8 \times 8} = \dfrac{5^2}{8^2} = \Big(\dfrac{5}{8}\Big)^2$

Hence, $\dfrac{25}{64} = \Big(\dfrac{5}{8}\Big)^2$.

(ii) Solving,

$-\dfrac{125}{216} = \dfrac{(-5) \times (-5) \times (-5)}{6 \times 6 \times 6} = \dfrac{(-5)^3}{6^3} = \Big(-\dfrac{5}{6}\Big)^3$

Hence, $-\dfrac{125}{216} = \Big(-\dfrac{5}{6}\Big)^3$.

(iii) Solving,

$-\dfrac{343}{729} = \dfrac{(-7) \times (-7) \times (-7)}{9 \times 9 \times 9} = \dfrac{(-7)^3}{9^3} = \Big(-\dfrac{7}{9}\Big)^3$

Hence, $-\dfrac{343}{729} = \Big(-\dfrac{7}{9}\Big)^3$.

Simplify the following:

(i) $\dfrac{(2^5)^2 \times 7^3}{8^3 \times 7}$

(ii) $\dfrac{25 \times 5^2 \times t^8}{10^3 \times t^4}$

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

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

Answer

(i) Solving,

$\dfrac{(2^5)^2 \times 7^3}{8^3 \times 7}\\[1em] = \dfrac{2^{5 \times 2} \times 7^3}{(2^3)^3 \times 7}\\[1em] = \dfrac{2^{10} \times 7^3}{2^9 \times 7^1}\\[1em] = 2^{10-9} \times 7^{3-1}\\[1em] = 2^1 \times 7^2\\[1em] = 2 \times 49\\[1em] = 98$

Hence, $\dfrac{(2^5)^2 \times 7^3}{8^3 \times 7} = 98$.

(ii) Solving,

$\dfrac{25 \times 5^2 \times t^8}{10^3 \times t^4}\\[1em] = \dfrac{5^2 \times 5^2 \times t^8}{(2 \times 5)^3 \times t^4}\\[1em] = \dfrac{5^{2+2} \times t^8}{2^3 \times 5^3 \times t^4}\\[1em] = \dfrac{5^4 \times t^8}{2^3 \times 5^3 \times t^4}\\[1em] = \dfrac{5^{4-3} \times t^{8-4}}{2^3}\\[1em] = \dfrac{5^1 \times t^4}{8}\\[1em] = \dfrac{5t^4}{8}$

Hence, $\dfrac{25 \times 5^2 \times t^8}{10^3 \times t^4} = \dfrac{5t^4}{8}$.

(iii) Solving,

$\dfrac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}\\[1em] = \dfrac{3^5 \times (2 \times 5)^5 \times 5^2}{5^7 \times (2 \times 3)^5}\\[1em] = \dfrac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5}\\[1em] = \dfrac{3^5 \times 2^5 \times 5^7}{5^7 \times 2^5 \times 3^5}\\[1em] = 3^{5-5} \times 2^{5-5} \times 5^{7-7}\\[1em] = 3^0 \times 2^0 \times 5^0\\[1em] = 1$

Hence, $\dfrac{3^5 \times 10^5 \times 25}{5^7 \times 6^5} = 1$.

(iv) Solving,

As we know, for any non-zero rational number a, $a^{-n} = \dfrac{1}{a^n}$.

$\Big(-\dfrac{3}{5}\Big)^{-3}\\[1em] = \Big(-\dfrac{5}{3}\Big)^3\\[1em] = \dfrac{(-5) \times (-5) \times (-5)}{3 \times 3 \times 3}\\[1em] = -\dfrac{125}{27}$

Hence, $\Big(-\dfrac{3}{5}\Big)^{-3} = -\dfrac{125}{27}$.

Simplify the following:

(i) $\Big(\dfrac{-1}{2}\Big)^5 \times 2^6 \times \Big(\dfrac{3}{4}\Big)^3$

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

Answer

(i) Solving,

$\Big(\dfrac{-1}{2}\Big)^5 \times 2^6 \times \Big(\dfrac{3}{4}\Big)^3\\[1em] = \dfrac{(-1)^5}{2^5} \times 2^6 \times \dfrac{3^3}{4^3}\\[1em] = \dfrac{-1}{2^5} \times 2^6 \times \dfrac{3^3}{(2^2)^3}\\[1em] = \dfrac{-1}{2^5} \times 2^6 \times \dfrac{3^3}{2^6}\\[1em] = -1 \times 2^{6-5-6} \times 3^3\\[1em] = -1 \times 2^{-5} \times 27\\[1em] = \dfrac{-27}{2^5}\\[1em] = -\dfrac{27}{32}$

Hence, $\Big(\dfrac{-1}{2}\Big)^5 \times 2^6 \times \Big(\dfrac{3}{4}\Big)^3 = -\dfrac{27}{32}$.

(ii) Solving,

$\Big[\Big(\dfrac{-3}{4}\Big)^3 - \Big(\dfrac{-5}{2}\Big)^3\Big] \times \Big(\dfrac{-2}{3}\Big)^4\\[1em] = \Big[\dfrac{(-3)^3}{4^3} - \dfrac{(-5)^3}{2^3}\Big] \times \dfrac{(-2)^4}{3^4}\\[1em] = \Big[\dfrac{-27}{64} - \dfrac{-125}{8}\Big] \times \dfrac{16}{81}\\[1em] = \Big[\dfrac{-27}{64} + \dfrac{125}{8}\Big] \times \dfrac{16}{81}$

LCM of 64 and 8 is 64

$= \Big[\dfrac{-27}{64} + \dfrac{125 \times 8}{8 \times 8}\Big] \times \dfrac{16}{81}\\[1em] = \Big[\dfrac{-27}{64} + \dfrac{1000}{64}\Big] \times \dfrac{16}{81}\\[1em] = \Big[\dfrac{-27 + 1000}{64}\Big] \times \dfrac{16}{81}\\[1em] = \dfrac{973}{64} \times \dfrac{16}{81}\\[1em] = \dfrac{973}{4 \times 81}\\[1em] = \dfrac{973}{324}\\[1em] = 3\dfrac{1}{324}$

Hence, $\Big[\Big(\dfrac{-3}{4}\Big)^3 - \Big(\dfrac{-5}{2}\Big)^3\Big] \times \Big(\dfrac{-2}{3}\Big)^4 = 3\dfrac{1}{324}$.

Simplify the following:

(i) $\Big(\dfrac{3}{2}\Big)^{-1} \div \Big(\dfrac{-2}{5}\Big)^{-1}$

(ii) $\Big[\Big\lbrace\Big(\dfrac{-1}{4}\Big)^2\Big\rbrace^{-1}\Big]^{-2}$

Answer

(i) Solving,

As we know, for any non-zero rational number a, $a^{-1} = \dfrac{1}{a}$.

$\Big(\dfrac{3}{2}\Big)^{-1} ÷ \Big(\dfrac{-2}{5}\Big)^{-1}\\[1em] = \dfrac{2}{3} ÷ \dfrac{-5}{2}\\[1em] = \dfrac{2}{3} \times \dfrac{2}{-5}\\[1em] = \dfrac{2 \times 2}{3 \times (-5)}\\[1em] = \dfrac{4}{-15}\\[1em] = -\dfrac{4}{15}$

Hence, $\Big(\dfrac{3}{2}\Big)^{-1} ÷ \Big(\dfrac{-2}{5}\Big)^{-1} = -\dfrac{4}{15}$.

(ii) Solving,

$\Big[\Big\lbrace\Big(\dfrac{-1}{4}\Big)^2\Big\rbrace^{-1}\Big]^{-2}\\[1em] = \Big[\Big(\dfrac{-1}{4}\Big)^{2 \times (-1)}\Big]^{-2}\\[1em] = \Big[\Big(\dfrac{-1}{4}\Big)^{-2}\Big]^{-2}\\[1em] = \Big(\dfrac{-1}{4}\Big)^{(-2) \times (-2)}\\[1em] = \Big(\dfrac{-1}{4}\Big)^4\\[1em] = \dfrac{(-1)^4}{4^4}\\[1em] = \dfrac{1}{256}$

Hence, $\Big[\Big\lbrace\Big(\dfrac{-1}{4}\Big)^2\Big\rbrace^{-1}\Big]^{-2} = \dfrac{1}{256}$.

Simplify: $\Big(\dfrac{1}{3}\Big)^{-2} + \Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{5}\Big)^{-2} - \Big(\dfrac{1}{6}\Big)^{-2}$

Answer

As we know, for any non-zero rational number a, $a^{-n} = \dfrac{1}{a^n}$.

$\Big(\dfrac{1}{3}\Big)^{-2} + \Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{5}\Big)^{-2} - \Big(\dfrac{1}{6}\Big)^{-2}\\[1em] = \Big(\dfrac{3}{1}\Big)^2 + \Big(\dfrac{4}{1}\Big)^2 + \Big(\dfrac{5}{1}\Big)^2 - \Big(\dfrac{6}{1}\Big)^2\\[1em] = 3^2 + 4^2 + 5^2 - 6^2\\[1em] = 9 + 16 + 25 - 36\\[1em] = 50 - 36\\[1em] = 14$

Hence, $\Big(\dfrac{1}{3}\Big)^{-2} + \Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{5}\Big)^{-2} - \Big(\dfrac{1}{6}\Big)^{-2} = 14$.

Express each of the following as a product of prime factors in the exponential form:

(i) 108 × 192

(ii) 729 × 64

(iii) 384 × 147

Answer

(i) By prime factorisation:

$\begin{array}{l|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

⇒ 108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³

⇒ 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁶ × 3¹.

So,

⇒ 108 × 192

⇒ (2² × 3³) × (2⁶ × 3¹)

⇒ 2²⁺⁶ × 3³⁺¹

⇒ 2⁸ × 3⁴.

Hence, 108 × 192 = 2⁸ × 3⁴.

(ii) By prime factorisation:

$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

⇒ 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶.

So,

⇒ 729 × 64 = 3⁶ × 2⁶.

Hence, 729 × 64 = 3⁶ × 2⁶.

(iii) By prime factorisation:

$\begin{array}{l|r} 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

⇒ 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁷ × 3¹

⇒ 147 = 3 × 7 × 7 = 3¹ × 7².

So,

⇒ 384 × 147

⇒ (2⁷ × 3¹) × (3¹ × 7²)

⇒ 2⁷ × 3¹⁺¹ × 7²

⇒ 2⁷ × 3² × 7².

Hence, 384 × 147 = 2⁷ × 3² × 7².

Simplify and write the following in the exponential form:

(i) 3³ × 2² + 2² × 5⁰

(ii) 9² + 11² - 2² × 3 × 17⁰

Answer

(i) Solving,

⇒ 3³ × 2² + 2² × 5⁰

⇒ (3 × 3 × 3) × (2 × 2) + (2 × 2) × 1

⇒ 27 × 4 + 4 × 1

⇒ 108 + 4

⇒ 112

⇒ 2 × 2 × 2 × 2 × 7

⇒ 2⁴ × 7¹.

Hence, 3³ × 2² + 2² × 5⁰ = 2⁴ × 7¹.

(ii) Solving,

⇒ 9² + 11² - 2² × 3 × 17⁰

⇒ (9 × 9) + (11 × 11) - (2 × 2) × 3 × 1

⇒ 81 + 121 - 4 × 3

⇒ 81 + 121 - 12

⇒ 190

⇒ 2 × 5 × 19

⇒ 2¹ × 5¹ × 19¹.

Hence, 9² + 11² - 2² × 3 × 17⁰ = 2¹ × 5¹ × 19¹.

(i) By what number should we multiply 3⁴ so that the product is 3⁷?

(ii) By what number should we multiply (-6)^-1 so that the product is 10^-1?

Answer

(i) Let the required number be x. Then,

$3^4 \times x = 3^7\\[1em] \Rightarrow x = \dfrac{3^7}{3^4}\\[1em] \Rightarrow x = 3^{7-4}\\[1em] \Rightarrow x = 3^3\\[1em] \Rightarrow x = 27$

Hence, 3⁴ should be multiplied by 27 to get 3⁷.

(ii) Let the required number be x. Then,

$(-6)^{-1} \times x = 10^{-1}\\[1em] \Rightarrow \dfrac{-1}{6} \times x = \dfrac{1}{10}\\[1em] \Rightarrow x = \dfrac{1}{10} \times \dfrac{6}{-1}\\[1em] \Rightarrow x = \dfrac{6}{-10}\\[1em] \Rightarrow x = -\dfrac{3}{5}$

Hence, (-6)^-1 should be multiplied by $-\dfrac{3}{5}$ to get 10^-1.

If $\Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{13}{12}\Big)^{-8} = \Big(\dfrac{12}{13}\Big)^{2x}$, then find the value of x.

Answer

$\Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{13}{12}\Big)^{-8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^4 \times \Big(\dfrac{12}{13}\Big)^{8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^{4+8} = \Big(\dfrac{12}{13}\Big)^{2x}\\[1em] \Rightarrow \Big(\dfrac{12}{13}\Big)^{12} = \Big(\dfrac{12}{13}\Big)^{2x}$

Using a^m = aⁿ ⇒ m = n

$\Rightarrow 2x = 12\\[1em] \Rightarrow x = \dfrac{12}{2}\\[1em] \Rightarrow x = 6$

Hence, x = 6.

If (-3)^x-1 = -243, then find the value of (-7)^x-6

Answer

⇒ (-3)^x-1 = -243

⇒ (-3)^x-1 = (-3) × (-3) × (-3) × (-3) × (-3)

⇒ (-3)^x-1 = (-3)⁵

⇒ x - 1 = 5

⇒ x = 6.

Now,

⇒ (-7)^x-6 = (-7)^6-6 = (-7)⁰ = 1.

Hence, (-7)^x-6 = 1.

Write the following numbers in the standard form (or scientific notation):

(i) 530.7

(ii) 3908.78

(iii) 39087.8

(iv) 2.35

(v) 3,43,000

(vi) 70,00,000

(vii) 3,18,65,00,000

(viii) 893,000,000

(ix) 70,040,000,000

Answer

To write a number in standard form, move the decimal point to the left till just one non-zero digit is left of it, then multiply by 10ⁿ, where n is the number of places moved.

(i) 530.7 = 5.307 × 10²

(ii) 3908.78 = 3.90878 × 10³

(iii) 39087.8 = 3.90878 × 10⁴

(iv) 2.35 = 2.35 × 10⁰

(v) 3,43,000 = 3.43 × 10⁵

(vi) 70,00,000 = 7.0 × 10⁶

(vii) 3,18,65,00,000 = 3.1865 × 10⁹

(viii) 893,000,000 = 8.93 × 10⁸

(ix) 70,040,000,000 = 7.004 × 10¹⁰

Write the following numbers in usual decimal notation:

(i) 4.7 × 10³

(ii) 1.205 × 10⁵

(iii) 1.234 × 10⁶

(iv) 4.87 × 10⁷

(v) 6.05 × 10⁸

(vi) 9.083 × 10¹¹

Answer

To write in usual form, move the decimal point to the right by the number of places given by the exponent of 10, adding trailing zeros as required.

(i) 4.7 × 10³ = 4700

(ii) 1.205 × 10⁵ = 120500

(iii) 1.234 × 10⁶ = 12,34,000

(iv) 4.87 × 10⁷ = 4,87,00,000

(v) 6.05 × 10⁸ = 60,50,00,000

(vi) 9.083 × 10¹¹ = 9,08,30,00,00,000

Express the numbers appearing in the following statements in scientific notation (or standard form):

(i) The distance between the earth and the moon is 384,000,000 m.

(ii) The diameter of the sun is 1,400,000,000 m.

(iii) The universe is estimated to be about 12,000,000,000 years old.

(iv) In a galaxy there are on an average 100,000,000,000 stars.

Answer

(i) Distance between the earth and the moon = 384,000,000 m = 3.84 × 10⁸ m

(ii) Diameter of the sun = 1,400,000,000 m = 1.4 × 10⁹ m

(iii) Age of the universe = 12,000,000,000 years = 1.2 × 10¹⁰ years

(iv) Average number of stars in a galaxy = 100,000,000,000 = 1.0 × 10¹¹

Compare the following numbers:

(i) 4.3 × 10¹⁴; 3.01 × 10¹⁷

(ii) 1.439 × 10¹²; 1.4335 × 10¹²

Answer

(i) The given numbers are 4.3 × 10¹⁴ and 3.01 × 10¹⁷.

Comparing the powers of 10, we have 17 > 14.

The number with the greater power of 10 is greater.

Hence, 3.01 × 10¹⁷ > 4.3 × 10¹⁴.

(ii) The given numbers are 1.439 × 10¹² and 1.4335 × 10¹².

Here the powers of 10 are equal, so we compare the significands.

As 1.439 > 1.4335,

Hence, 1.439 × 10¹² > 1.4335 × 10¹².

Write the following numbers in the expanded exponential form:

(i) 279404

(ii) 3006194

(iii) 28061906

Answer

(i) Solving,

⇒ 279404

⇒ 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1

⇒ 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 4 × 10⁰.

Hence, 279404 = 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 4 × 10⁰.

(ii) Solving,

⇒ 3006194

⇒ 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1

⇒ 3 × 10⁶ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰.

Hence, 3006194 = 3 × 10⁶ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰.

(iii) Solving,

⇒ 28061906

⇒ 2 × 10000000 + 8 × 1000000 + 0 × 100000 + 6 × 10000 + 1 × 1000 + 9 × 100 + 0 × 10 + 6 × 1

⇒ 2 × 10⁷ + 8 × 10⁶ + 6 × 10⁴ + 1 × 10³ + 9 × 10² + 6 × 10⁰.

Hence, 28061906 = 2 × 10⁷ + 8 × 10⁶ + 6 × 10⁴ + 1 × 10³ + 9 × 10² + 6 × 10⁰.

Find the number from each of the expanded form:

(i) 3 × 10⁴ + 7 × 10² + 5 × 10⁰

(ii) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

(iii) 8 × 10⁷ + 3 × 10⁴ + 7 × 10³ + 5 × 10² + 8 × 10¹

Answer

(i) Solving,

⇒ 3 × 10⁴ + 7 × 10² + 5 × 10⁰

⇒ 3 × 10000 + 7 × 100 + 5 × 1

⇒ 30000 + 700 + 5

⇒ 30705.

Hence, the number is 30705.

(ii) Solving,

⇒ 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

⇒ 4 × 100000 + 5 × 1000 + 3 × 100 + 2 × 1

⇒ 400000 + 5000 + 300 + 2

⇒ 405302.

Hence, the number is 405302.

(iii) Solving,

⇒ 8 × 10⁷ + 3 × 10⁴ + 7 × 10³ + 5 × 10² + 8 × 10¹

⇒ 8 × 10000000 + 3 × 10000 + 7 × 1000 + 5 × 100 + 8 × 10

⇒ 80000000 + 30000 + 7000 + 500 + 80

⇒ 80037580.

Hence, the number is 80037580.

Fill in the blanks:

(i) In the expression (-5)⁹, exponent = ............... and base = ...............

(ii) If the base is $-\dfrac{3}{4}$ and exponent is 5, then exponential form is ...............

(iii) The expression (x²y⁵)³ in the simplest form is ...............

(iv) If (100)⁰ = 10ⁿ, then the value of n is ...............

(v) $\Big(-\dfrac{1}{2}\Big)^0 + (-2)^0$ = ...............

(vi) (-3)² × (-1)²⁰¹⁷ = ...............

(vii) (-3)⁸ ÷ (-3)⁵ = (-3)^...

(viii) 35070000 = 3.507 × 10^...

(ix) If (-2)ⁿ = -128, then n = ...............

Answer

(i) In (-5)⁹, exponent = 9 and base = -5

(ii) If base is $-\dfrac{3}{4}$ and exponent is 5, then exponential form is $\mathbf{\Big(-\dfrac{3}{4}\Big)^5}$

(iii) Solving,

⇒ (x² y⁵)³ = (x²)³ × (y⁵)³ = x^{2 × 3} × y^{5 × 3} = x⁶ y¹⁵.

So the simplest form is x⁶y¹⁵

(iv) (100)⁰ = 1 = 10⁰, so the value of n is 0

(v)

$\Big(-\dfrac{1}{2}\Big)^0 + (-2)^0 = 1 + 1 = 2$

So the value is 2

(vi) Solving,

⇒ (-3)² × (-1)²⁰¹⁷ = 9 × (-1) = -9.

So the value is -9

(vii) Solving,

⇒ (-3)⁸ ÷ (-3)⁵ = (-3)^8-5 = (-3)³.

So the blank is 3

(viii) 35070000 = 3.507 × 10⁷, so the blank is 7

(ix) (-2)ⁿ = -128 = (-2)⁷, so n = 7

State whether the following statements are true (T) or false (F):

(i) If a is a rational number then a^m × aⁿ = a^{m × n}

(ii) 2³ × 3² = 6⁵

(iii) The value of (-2)^-3 is $-\dfrac{1}{8}$

(iv) The value of the expression 2⁹ × 2⁹¹ - 2¹⁹ × 2⁸¹ is 1

(v) 5⁶ ÷ (-2)⁶ = $-\dfrac{5}{2}$

(vi) 5⁰ × 3⁰ = 8⁰

(vii) $\dfrac{2^3}{7} \lt \Big(\dfrac{2}{7}\Big)^3$

(viii) (10 + 10)⁴ = 10⁴ + 10⁴

(ix) x⁰ × x⁰ = x⁰ ÷ x⁰, where x is a non-zero rational number.

(x) 4⁹ is greater than 16³

(xi) x^m + x^m = x^2m, where x is a non-zero rational number and m is a positive integer.

(xii) $\Big(\dfrac{4}{3}\Big)^5 \times \Big(\dfrac{5}{7}\Big)^5 = \Big(\dfrac{4}{3}+\dfrac{5}{7}\Big)^5$

Answer

(i) False. By the law of exponents, a^m × aⁿ = a^{m + n}, not a^{m × n}.

(ii) False. 2³ × 3² = 8 × 9 = 72, whereas 6⁵ = 7776.

(iii) True. $(-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}$.

(iv) False. 2⁹ × 2⁹¹ - 2¹⁹ × 2⁸¹ = 2¹⁰⁰ - 2¹⁰⁰ = 0, not 1.

(v) False. $5^6 ÷ (-2)^6 = \dfrac{5^6}{2^6} = \Big(\dfrac{5}{2}\Big)^6$, which is positive, not $-\dfrac{5}{2}$.

(vi) True. 5⁰ × 3⁰ = 1 × 1 = 1 = 8⁰.

(vii) False. $\dfrac{2^3}{7} = \dfrac{8}{7}$ and $\Big(\dfrac{2}{7}\Big)^3 = \dfrac{8}{343}$. Since $\dfrac{8}{7} \gt \dfrac{8}{343}$, the statement is false.

(viii) False. (10 + 10)⁴ = 20⁴ = 160000, whereas 10⁴ + 10⁴ = 20000.

(ix) True. x⁰ × x⁰ = 1 × 1 = 1 and x⁰ ÷ x⁰ = 1 ÷ 1 = 1.

(x) True. 4⁹ = 262144 and 16³ = 4096, so 4⁹ > 16³.

(xi) False. x^m + x^m = 2x^m, which is not equal to x^2m.

(xii) False. Powers with the same exponent multiply the bases: $\Big(\dfrac{4}{3}\Big)^5 \times \Big(\dfrac{5}{7}\Big)^5 = \Big(\dfrac{4}{3} \times \dfrac{5}{7}\Big)^5$, not $\Big(\dfrac{4}{3} + \dfrac{5}{7}\Big)^5$.

a × a × a × b × b × b is equal to

1. a³b²

2. a²b³

3. (ab)³

4. a⁶b⁶

Answer

⇒ a × a × a × b × b × b

⇒ a³ × b³

⇒ (ab)³.

Hence, option 3 is the correct option.

(-2)³ × (-3)² is equal to

1. 6⁵

2. (-6)⁶

3. 72

4. -72

Answer

⇒ (-2)³ × (-3)²

⇒ [(-2) × (-2) × (-2)] × [(-3) × (-3)]

⇒ (-8) × 9

⇒ -72.

Hence, option 4 is the correct option.

The expression (pqr)³ is equal to

1. p³qr

2. pq³r

3. pqr³

4. p³q³r³

Answer

⇒ (pqr)³ = p³ × q³ × r³ = p³ q³ r³.

Hence, option 4 is the correct option.

$\Big(-\dfrac{3}{2}\Big)^{-1}$ is equal to

1. $\dfrac{2}{3}$

2. $-\dfrac{2}{3}$

3. $\dfrac{3}{2}$

4. $\dfrac{4}{9}$

Answer

As we know, for any non-zero rational number a, $a^{-1} = \dfrac{1}{a}$.

$\Big(-\dfrac{3}{2}\Big)^{-1} = -\dfrac{2}{3}$

Hence, option 2 is the correct option.

$\Big(-\dfrac{3}{4}\Big)^5$ is equal to

1. $\dfrac{81}{256}$

2. $-\dfrac{81}{256}$

3. $-\dfrac{243}{1024}$

4. $\dfrac{243}{1024}$

Answer

$\Big(-\dfrac{3}{4}\Big)^5\\[1em] = \dfrac{(-3)^5}{4^5}\\[1em] = \dfrac{(-3) \times (-3) \times (-3) \times (-3) \times (-3)}{4 \times 4 \times 4 \times 4 \times 4}\\[1em] = -\dfrac{243}{1024}$

Hence, option 3 is the correct option.

The value of (5³⁰ × 5²⁰) ÷ (5⁵)⁹ in the exponential form is

1. 5^-5

2. 5⁵

3. 5⁵⁰

4. 5⁹⁵

Answer

⇒ (5³⁰ × 5²⁰) ÷ (5⁵)⁹

⇒ 5³⁰⁺²⁰ ÷ 5^{5 × 9}

⇒ 5⁵⁰ ÷ 5⁴⁵

⇒ 5^50-45

⇒ 5⁵.

Hence, option 2 is the correct option.

The law $\Big(\dfrac{a}{b}\Big)^n = \dfrac{a^n}{b^n}$ does not hold when

1. a = 3, b = 2

2. a = -2, b = 3

3. n = 0

4. b = 0

Answer

The expression $\Big(\dfrac{a}{b}\Big)^n = \dfrac{a^n}{b^n}$ involves division by b, which is not defined when b = 0.

Hence, option 4 is the correct option.

The value of the expression $\dfrac{(-1)^{101} \times (8)^5}{(4)^7}$ is equal to

1. 2

2. -2

3. $\dfrac{1}{16}$

4. $-\dfrac{1}{16}$

Answer

$\dfrac{(-1)^{101} \times (8)^5}{(4)^7}\\[1em] = \dfrac{(-1) \times (2^3)^5}{(2^2)^7}\\[1em] = \dfrac{(-1) \times 2^{15}}{2^{14}}\\[1em] = (-1) \times 2^{15-14}\\[1em] = (-1) \times 2^1\\[1em] = -2$

Hence, option 2 is the correct option.

The value of $\dfrac{10^{22} + 10^{20}}{10^{20}}$ is

1. 10

2. 101

3. 10²²

4. 10⁴²

Answer

$\dfrac{10^{22} + 10^{20}}{10^{20}}\\[1em] = \dfrac{10^{20}(10^2 + 1)}{10^{20}}\\[1em] = 10^2 + 1\\[1em] = 100 + 1\\[1em] = 101$

Hence, option 2 is the correct option.

The value of 5^-1 - 6^-1 is

1. $\dfrac{1}{30}$

2. $-\dfrac{1}{30}$

3. 30

4. -30

Answer

$5^{-1} - 6^{-1}\\[1em] = \dfrac{1}{5} - \dfrac{1}{6}$

LCM of 5 and 6 is 30

$= \dfrac{1 \times 6}{5 \times 6} - \dfrac{1 \times 5}{6 \times 5}\\[1em] = \dfrac{6}{30} - \dfrac{5}{30}\\[1em] = \dfrac{6 - 5}{30}\\[1em] = \dfrac{1}{30}$

Hence, option 1 is the correct option.

The value of (6^-1 - 8^-1)^-1 is

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

2. -2

3. $\dfrac{1}{24}$

4. 24

Answer

Solving,

$\Rightarrow (6^{-1} - 8^{-1})^{-1}\\[1em] = \Big(\dfrac{1}{6} - \dfrac{1}{8}\Big)^{-1}$

LCM of 6 and 8 is 24

$= \Big(\dfrac{1 \times 4}{6 \times 4} - \dfrac{1 \times 3}{8 \times 3}\Big)^{-1}\\[1em] = \Big(\dfrac{4}{24} - \dfrac{3}{24}\Big)^{-1}\\[1em] = \Big(\dfrac{4 - 3}{24}\Big)^{-1}\\[1em] = \Big(\dfrac{1}{24}\Big)^{-1}\\[1em] = 24$

Hence, option 4 is the correct option.

The value of $\Big(\Big(\dfrac{1}{3}\Big)^{-2}+\Big(\dfrac{1}{4}\Big)^{-2}\Big) ÷ \Big(\dfrac{1}{5}\Big)^{-2}$ is

1. 0

2. -1

3. 1

4. $\dfrac{7}{5}$

Answer

As we know, for any non-zero rational number a, $a^{-n} = \dfrac{1}{a^n}$.

$\Big(\Big(\dfrac{1}{3}\Big)^{-2}+\Big(\dfrac{1}{4}\Big)^{-2}\Big) ÷ \Big(\dfrac{1}{5}\Big)^{-2}\\[1em] = (3^2 + 4^2) ÷ 5^2\\[1em] = (9 + 16) ÷ 25\\[1em] = 25 ÷ 25\\[1em] = 1$

Hence, option 3 is the correct option.

If 2³ + 1³ = 3^x, then the value of x is

1. 0

2. 1

3. 2

4. 3

Answer

⇒ 2³ + 1³ = 3^x

⇒ 8 + 1 = 3^x

⇒ 9 = 3^x

⇒ 3² = 3^x

⇒ x = 2.

Hence, option 3 is the correct option.

The standard form of 751.65 is

1. 7.5165 × 10²

2. 75.165 × 10¹

3. 7.5165 × 10⁴

4. 7.51 × 10²

Answer

Moving the decimal point two places to the left,

⇒ 751.65 = 7.5165 × 10².

Hence, option 1 is the correct option.

The usual form of 5.658 × 10⁵ is

1. 5658

2. 56580

3. 565800

4. 5658000

Answer

Moving the decimal point five places to the right,

⇒ 5.658 × 10⁵ = 565800.

Hence, option 3 is the correct option.

Which of the following numbers is in the standard form?

1. 26.57 × 10⁴

2. 2.657 × 10⁵

3. 265.7 × 10³

4. 0.2657 × 10⁶

Answer

A number is in standard form k × 10ⁿ only when 1 ≤ k < 10. Only 2.657 satisfies this condition.

Hence, option 2 is the correct option.

Statement I: $\Big(\dfrac{a}{b}\Big)^n$ is not defined if n = 0

Statement II: $\Big(\dfrac{a}{b}\Big)^n$ is not defined if b = 0

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

Statement I: When n = 0, $\Big(\dfrac{a}{b}\Big)^0 = 1$, which is well defined. So Statement I is false.