Exponents and Powers

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

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

(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,

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

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

(v) Solving,

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

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

(vi) Solving,

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

Hence, (-2)⁹ = -512.

(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

(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.

(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.

(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.

(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.

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.

(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.

(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,

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

Hence, x = 5.

(vi) Solving,

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

Hence, x = 5.

(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.

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,

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

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

(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}.

(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.

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.

(i) Solving,

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

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

(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.

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)

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

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

(i) Solving,

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

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

(ii) Solving,

$$\begin{aligned}\frac{(3^2)^3 \times (-2)^5}{(-2)^3}\\[1em] = \frac{3^{2 \times 3} \times (-2)^5}{(-2)^3}\\[1em] = 3^6 \times (-2)^{5-3}\\[1em] = 3^6 \times (-2)^2\\[1em] = 3^6 \times 2^2\end{aligned}$$

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

(iii) Solving,

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

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

(iv) Solving,

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

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

(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.

(i) Solving,

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

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

(ii) Solving,

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

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

(iii) Solving,

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

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

(i) Solving,

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

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

(ii) Solving,

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

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

(iii) Solving,

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

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

(iv) Solving,

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

$$\begin{aligned}\Big(-\frac{3}{5}\Big)^{-3}\\[1em] = \Big(-\frac{5}{3}\Big)^3\\[1em] = \frac{(-5) \times (-5) \times (-5)}{3 \times 3 \times 3}\\[1em] = -\frac{125}{27}\end{aligned}$$

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

(i) Solving,

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

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

(ii) Solving,

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

LCM of 64 and 8 is 64

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

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

(i) Solving,

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

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

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

(ii) Solving,

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

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

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

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

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

(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.

(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) Let the required number be x. Then,

$$\begin{aligned}3^4 \times x = 3^7\\[1em] \Rightarrow x = \frac{3^7}{3^4}\\[1em] \Rightarrow x = 3^{7-4}\\[1em] \Rightarrow x = 3^3\\[1em] \Rightarrow x = 27\end{aligned}$$

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

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

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

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

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

Using a^m = aⁿ ⇒ m = n

$$\begin{aligned}\Rightarrow 2x = 12\\[1em] \Rightarrow x = \frac{12}{2}\\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, x = 6.

⇒ (-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.

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

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

(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

(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.

(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.

(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.

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

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

(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)

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

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

(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. $(-2)^{-3} = \frac{1}{(-2)^3} = \frac{1}{-8} = -\frac{1}{8}$.

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

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

(vi) True. 5⁰ × 3⁰ = 1 × 1 = 1 = 8⁰.

(vii) False. $\frac{2^3}{7} = \frac{8}{7}$ and $\Big(\frac{2}{7}\Big)^3 = \frac{8}{343}$. Since $\frac{8}{7} \gt \frac{8}{343}$, 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: $\Big(\frac{4}{3}\Big)^5 \times \Big(\frac{5}{7}\Big)^5 = \Big(\frac{4}{3} \times \frac{5}{7}\Big)^5$, not $\Big(\frac{4}{3} + \frac{5}{7}\Big)^5$.

⇒ a × a × a × b × b × b

⇒ a³ × b³

⇒ (ab)³.

Hence, option 3 is the correct option.

⇒ (-2)³ × (-3)²

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

⇒ (-8) × 9

⇒ -72.

Hence, option 4 is the correct option.

⇒ (pqr)³ = p³ × q³ × r³ = p³ q³ r³.

Hence, option 4 is the correct option.

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

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

Hence, option 2 is the correct option.

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

Hence, option 3 is the correct option.

⇒ (5³⁰ × 5²⁰) ÷ (5⁵)⁹

⇒ 5³⁰⁺²⁰ ÷ 5^{5 × 9}

⇒ 5⁵⁰ ÷ 5⁴⁵

⇒ 5^50-45

⇒ 5⁵.

Hence, option 2 is the correct option.

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

Hence, option 4 is the correct option.

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

Hence, option 2 is the correct option.

$$\begin{aligned}\frac{10^{22} + 10^{20}}{10^{20}}\\[1em] = \frac{10^{20}(10^2 + 1)}{10^{20}}\\[1em] = 10^2 + 1\\[1em] = 100 + 1\\[1em] = 101\end{aligned}$$

Hence, option 2 is the correct option.

$$\begin{aligned}5^{-1} - 6^{-1}\\[1em] = \frac{1}{5} - \frac{1}{6}\end{aligned}$$

LCM of 5 and 6 is 30

$$\begin{aligned}= \frac{1 \times 6}{5 \times 6} - \frac{1 \times 5}{6 \times 5}\\[1em] = \frac{6}{30} - \frac{5}{30}\\[1em] = \frac{6 - 5}{30}\\[1em] = \frac{1}{30}\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (6^{-1} - 8^{-1})^{-1}\\[1em] = \Big(\frac{1}{6} - \frac{1}{8}\Big)^{-1}\end{aligned}$$

LCM of 6 and 8 is 24

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

Hence, option 4 is the correct option.

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

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

Hence, option 3 is the correct option.

⇒ 2³ + 1³ = 3^x

⇒ 8 + 1 = 3^x

⇒ 9 = 3^x

⇒ 3² = 3^x

⇒ x = 2.

Hence, option 3 is the correct option.

Moving the decimal point two places to the left,

⇒ 751.65 = 7.5165 × 10².

Hence, option 1 is the correct option.

Moving the decimal point five places to the right,

⇒ 5.658 × 10⁵ = 565800.

Hence, option 3 is the correct option.

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

Statement II: When b = 0, the base $\frac{a}{b}$ itself is not defined (division by zero). So Statement II is true.

Hence, option 2 is the correct option.

Statement I: Since 300 is even,

$$\begin{aligned}(-3)^{300} = 3^{300};\text{(a very large positive number)}\\[1em] (-3)^{-300} = \frac{1}{(-3)^{300}} = \frac{1}{3^{300}};\text{(a very small positive number)}\end{aligned}$$

So (-3)^-300 < (-3)³⁰⁰, which means the claim (-3)^-300 > (-3)³⁰⁰ is false.

Statement II: For any non-zero rational number a, $a^{-1} = \frac{1}{a}$, which is indeed the reciprocal of a. So Statement II is true.

Hence, option 2 is the correct option.

Statement I:

⇒ 32⁴ = (2⁵)⁴ = 2²⁰

⇒ 16⁵ = (2⁴)⁵ = 2²⁰.

So 32⁴ = 16⁵, which means 32⁴ < 16⁵ is false.

Statement II: The claim x^a > y^b whenever x > y is not true in general (for example 3¹ = 3 is not greater than 2⁵ = 32). So Statement II is false.

Hence, option 4 is the correct option.

(i) Solving,

⇒ (-3)³ × 5²

⇒ [(-3) × (-3) × (-3)] × (5 × 5)

⇒ (-27) × 25

⇒ -675.

Hence, (-3)³ × 5² = -675.

(ii)

As 501 is an odd natural number, (-1)⁵⁰¹ = -1.

⇒ (-1)⁵⁰¹ × [(27)⁴ ÷ (9)⁵]

⇒ -1 × [(3³)⁴ ÷ (3²)⁵]

⇒ -1 × [3¹² ÷ 3¹⁰]

⇒ -1 × 3^12-10

⇒ -1 × 3²

⇒ -1 × 9

⇒ -9.

Hence, (-1)⁵⁰¹ × [(27)⁴ ÷ (9)⁵] = -9.

(iii)

$$\begin{aligned}\Big(-3\frac{1}{2}\Big)^3\\[1em] = \Big(-\frac{7}{2}\Big)^3\\[1em] = \frac{(-7) \times (-7) \times (-7)}{2 \times 2 \times 2}\\[1em] = -\frac{343}{8}\\[1em] = -42\frac{7}{8}\end{aligned}$$

Hence, $\Big(-3\frac{1}{2}\Big)^3 = -42\frac{7}{8}$.

(i) Solving,

$$\begin{aligned}\frac{7^3 \times 11^4 \times 13^0}{7^2 \times 11^2}\\[1em] = \frac{7^3 \times 11^4 \times 1}{7^2 \times 11^2}\\[1em] = 7^{3-2} \times 11^{4-2}\\[1em] = 7^1 \times 11^2\\[1em] = 7 \times 121\\[1em] = 847\end{aligned}$$

Hence, $\frac{7^3 \times 11^4 \times 13^0}{7^2 \times 11^2} = 847$.

(ii) Solving,

$$\begin{aligned}\frac{(-2)^3 \times (3x)^2 \times (-xy^3)}{3x^2 y}\\[1em] = \frac{(-8) \times 9x^2 \times (-xy^3)}{3x^2 y}\\[1em] = \frac{(-8) \times 9 \times (-1) \times x^2 \times x \times y^3}{3x^2 y}\\[1em] = \frac{72 \times x^3 \times y^3}{3 x^2 y}\\[1em] = \frac{72}{3} \times x^{3-2} \times y^{3-1}\\[1em] = 24 \times x^1 \times y^2\\[1em] = 24xy^2\end{aligned}$$

Hence, $\frac{(-2)^3 \times (3x)^2 \times (-xy^3)}{3x^2 y} = 24xy^2$.

(iii) Solving,

$$\begin{aligned}\frac{((-5)^3)^4 \times 8^2}{4^3 \times (25)^5}\\[1em] = \frac{(-5)^{3 \times 4} \times (2^3)^2}{(2^2)^3 \times (5^2)^5}\\[1em] = \frac{(-5)^{12} \times 2^6}{2^6 \times 5^{10}}\\[1em] = \frac{5^{12} \times 2^6}{2^6 \times 5^{10}}\\[1em] = 5^{12-10} \times 2^{6-6}\\[1em] = 5^2 \times 2^0\\[1em] = 25 \times 1\\[1em] = 25\end{aligned}$$

Hence, $\frac{((-5)^3)^4 \times 8^2}{4^3 \times (25)^5} = 25$.

(i) Solving,

$$\begin{aligned}\frac{(-3)^5 \times 8^3 \times 2^5}{3^2 \times 4^4}\\[1em] = \frac{(-3)^5 \times (2^3)^3 \times 2^5}{3^2 \times (2^2)^4}\\[1em] = \frac{(-1)^5 \times 3^5 \times 2^9 \times 2^5}{3^2 \times 2^8}\\[1em] = \frac{-3^5 \times 2^{9+5}}{3^2 \times 2^8}\\[1em] = \frac{-3^5 \times 2^{14}}{3^2 \times 2^8}\\[1em] = -3^{5-2} \times 2^{14-8}\\[1em] = -3^3 \times 2^6\\[1em] = (-3)^3 \times 2^6\end{aligned}$$

Hence, $\frac{(-3)^5 \times 8^3 \times 2^5}{3^2 \times 4^4} = (-3)^3 \times 2^6$.

(ii) Solving,

$$\begin{aligned}\frac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2}\\[1em] = \frac{(3^2)^8 \times x^{2 \times 5}}{(3^3)^4 \times x^{3 \times 2}}\\[1em] = \frac{3^{16} \times x^{10}}{3^{12} \times x^6}\\[1em] = 3^{16-12} \times x^{10-6}\\[1em] = 3^4 \times x^4\\[1em] = (3x)^4\end{aligned}$$

Hence, $\frac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2} = (3x)^4$.

(iii) Solving,

$$\begin{aligned}\frac{3^2 \times 7^8 \times 13^6}{21^2 \times 91^3}\\[1em] = \frac{3^2 \times 7^8 \times 13^6}{(3 \times 7)^2 \times (7 \times 13)^3}\\[1em] = \frac{3^2 \times 7^8 \times 13^6}{3^2 \times 7^2 \times 7^3 \times 13^3}\\[1em] = \frac{3^2 \times 7^8 \times 13^6}{3^2 \times 7^5 \times 13^3}\\[1em] = 3^{2-2} \times 7^{8-5} \times 13^{6-3}\\[1em] = 3^0 \times 7^3 \times 13^3\\[1em] = 7^3 \times 13^3\\[1em] = (7 \times 13)^3\\[1em] = 91^3\end{aligned}$$

Hence, $\frac{3^2 \times 7^8 \times 13^6}{21^2 \times 91^3} = 91^3$.

Solving,

$$\begin{aligned}\Big(-\frac{3}{5}\Big)^x = -\frac{27}{125}\\[1em] \Rightarrow \Big(-\frac{3}{5}\Big)^x = \frac{(-3) \times (-3) \times (-3)}{5 \times 5 \times 5}\\[1em] \Rightarrow \Big(-\frac{3}{5}\Big)^x = \Big(-\frac{3}{5}\Big)^3\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, x = 3.

(i) By prime factorisation of 24000:

$$\begin{array}{l|r} 2 & 24000 \\ \hline 2 & 12000 \\ \hline 2 & 6000 \\ \hline 2 & 3000 \\ \hline 2 & 1500 \\ \hline 2 & 750 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

⇒ 24000 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5

⇒ 2⁶ × 3¹ × 5³.

Hence, 24000 = 2⁶ × 3¹ × 5³.

(ii) By prime factorisation of 12600:

$$\begin{array}{l|r} 2 & 12600 \\ \hline 2 & 6300 \\ \hline 2 & 3150 \\ \hline 3 & 1575 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

⇒ 12600 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 7

⇒ 2³ × 3² × 5² × 7¹.

Hence, 12600 = 2³ × 3² × 5² × 7¹.

(iii) By prime factorisation of 14157:

$$\begin{array}{l|r} 3 & 14157 \\ \hline 3 & 4719 \\ \hline 11 & 1573 \\ \hline 11 & 143 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

⇒ 14157 = 3 × 3 × 11 × 11 × 13

⇒ 3² × 11² × 13¹.

Hence, 14157 = 3² × 11² × 13¹.

(i) Volume of water on the earth = 1,353,000,000 cubic km = 1.353 × 10⁹ cubic km

(ii) Population of India = 1,429,000,000 = 1.429 × 10⁹

(i) The given numbers are 5.976 × 10²⁴ and 8.689 × 10²³.

Comparing the powers of 10, we have 24 > 23.

The number with the greater power of 10 is greater.

Hence, 5.976 × 10²⁴ > 8.689 × 10²³.

(ii) The given numbers are 3.7662 × 10¹⁷ and 3.7671 × 10¹⁷.

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

As 3.7671 > 3.7662,

Hence, 3.7671 × 10¹⁷ > 3.7662 × 10¹⁷.

$$\begin{aligned}\Big(\frac{5}{12}\Big)^8 \times \Big(\frac{12}{5}\Big)^{-4} = \Big(\frac{125}{1728}\Big)^x\\[1em] \Rightarrow \Big(\frac{5}{12}\Big)^8 \times \Big(\frac{5}{12}\Big)^{4} = \Big(\frac{125}{1728}\Big)^x\\[1em] \Rightarrow \Big(\frac{5}{12}\Big)^{8+4} = \Big(\frac{5^3}{12^3}\Big)^x\\[1em] \Rightarrow \Big(\frac{5}{12}\Big)^{12} = \Big(\Big(\frac{5}{12}\Big)^3\Big)^x\\[1em] \Rightarrow \Big(\frac{5}{12}\Big)^{12} = \Big(\frac{5}{12}\Big)^{3x}\end{aligned}$$

Using a^m = aⁿ ⇒ m = n

$$\begin{aligned}\Rightarrow 3x = 12\\[1em] \Rightarrow x = \frac{12}{3}\\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, x = 4.

⇒ 27^2x-1 = (243)³

⇒ (3³)^2x-1 = (3⁵)³

⇒ 3^3(2x-1) = 3¹⁵

⇒ 3^6x-3 = 3¹⁵.

Using a^m = aⁿ ⇒ m = n

⇒ 6x - 3 = 15

⇒ 6x = 18

⇒ x = 3.

Now,

⇒ (-5)^2x-3 = (-5)^{2 × 3 - 3} = (-5)^6-3 = (-5)³ = -125.

Hence, (-5)^2x-3 = -125.

Using the order of operations (of, then ÷, then -),

$$\begin{aligned}\Rightarrow x = \frac{1}{6} - \frac{5}{3} \text{ of } \frac{2}{-5} ÷ \frac{4}{9}\\[1em] = \frac{1}{6} - \Big(\frac{5}{3} \times \frac{2}{-5}\Big) ÷ \frac{4}{9}\\[1em] = \frac{1}{6} - \Big(\frac{10}{-15}\Big) ÷ \frac{4}{9}\\[1em] = \frac{1}{6} - \Big(-\frac{2}{3}\Big) ÷ \frac{4}{9}\\[1em] = \frac{1}{6} - \Big(-\frac{2}{3} \times \frac{9}{4}\Big)\\[1em] = \frac{1}{6} - \Big(-\frac{18}{12}\Big)\\[1em] = \frac{1}{6} - \Big(-\frac{3}{2}\Big)\\[1em] = \frac{1}{6} + \frac{3}{2}\end{aligned}$$

LCM of 6 and 2 is 6

$$\begin{aligned}= \frac{1}{6} + \frac{3 \times 3}{2 \times 3}\\[1em] = \frac{1}{6} + \frac{9}{6}\\[1em] = \frac{1 + 9}{6}\\[1em] = \frac{10}{6}\\[1em] = \frac{5}{3}\end{aligned}$$

Now,

$$\begin{aligned}\Rightarrow x^3 - 3\\[1em] = \Big(\frac{5}{3}\Big)^3 - 3\\[1em] = \frac{125}{27} - 3\\[1em] = \frac{125}{27} - \frac{3 \times 27}{27}\\[1em] = \frac{125 - 81}{27}\\[1em] = \frac{44}{27}\\[1em] = 1\frac{17}{27}\end{aligned}$$

Hence, x³ - 3 = $1\frac{17}{27}$.