Exponents

(i) Solving,

⇒ 6² = 6 × 6 = 36.

Hence, 6² = 36.

(ii) Solving,

⇒ 7³ = 7 × 7 × 7 = 343.

Hence, 7³ = 343.

(iii) Solving,

⇒ 4⁴ = 4 × 4 × 4 × 4 = 256.

Hence, 4⁴ = 256.

(iv) Solving,

⇒ 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125.

Hence, 5⁵ = 3125.

(v) Solving,

⇒ 8³ = 8 × 8 × 8 = 512.

Hence, 8³ = 512.

(vi) Solving,

⇒ 7⁵ = 7 × 7 × 7 × 7 × 7 = 16807.

Hence, 7⁵ = 16807.

(i) Solving,

⇒ 2³ × 4² = (2 × 2 × 2) × (4 × 4) = 8 × 16 = 128.

Hence, 2³ × 4² = 128.

(ii) Solving,

⇒ 2³ × 5² = (2 × 2 × 2) × (5 × 5) = 8 × 25 = 200.

Hence, 2³ × 5² = 200.

(iii) Solving,

⇒ 3³ × 5² = (3 × 3 × 3) × (5 × 5) = 27 × 25 = 675.

Hence, 3³ × 5² = 675.

(iv) Solving,

⇒ 2² × 3³ = (2 × 2) × (3 × 3 × 3) = 4 × 27 = 108.

Hence, 2² × 3³ = 108.

(v) Solving,

⇒ 3² × 5³ = (3 × 3) × (5 × 5 × 5) = 9 × 125 = 1125.

Hence, 3² × 5³ = 1125.

(vi) Solving,

⇒ 5³ × 2⁴ = (5 × 5 × 5) × (2 × 2 × 2 × 2) = 125 × 16 = 2000.

Hence, 5³ × 2⁴ = 2000.

(vii) Solving,

⇒ 3² × 4² = (3 × 3) × (4 × 4) = 9 × 16 = 144.

Hence, 3² × 4² = 144.

(viii) Solving,

⇒ (4 × 3)³ = 12³ = 12 × 12 × 12 = 1728.

Hence, (4 × 3)³ = 1728.

(ix) Solving,

⇒ (5 × 4)² = 20² = 20 × 20 = 400.

Hence, (5 × 4)² = 400.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{3}{4}\right)^4 \\[1em] = \frac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4} \\[1em] = \frac{81}{256}\end{aligned}$$

Hence, $\left(\frac{3}{4}\right)^4 = \frac{81}{256}$.

Solving,

$$\begin{aligned}\Rightarrow \left(-\frac{5}{6}\right)^5 \\[1em] = \frac{(-5) \times (-5) \times (-5) \times (-5) \times (-5)}{6 \times 6 \times 6 \times 6 \times 6} \\[1em] = -\frac{3125}{7776}\end{aligned}$$

Hence, $\left(-\frac{5}{6}\right)^5 = -\frac{3125}{7776}$.

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{2}{3}\right)^3 \times \left(\frac{3}{4}\right)^2\\[1em] = \frac{2 \times 2 \times 2}{3 \times 3 \times 3} \times \frac{3 \times 3}{4 \times 4}\\[1em] = \frac{2 \times 2 \times 2 \times 3 \times 3}{3 \times 3 \times 3 \times 4 \times 4}\\[1em] = \frac{8}{3 \times 16}\\[1em] = \frac{8}{48}\\[1em] = \frac{1}{6}\end{aligned}$$

Hence, $\left(\frac{2}{3}\right)^3 \times \left(\frac{3}{4}\right)^2 = \frac{1}{6}$.

Solving,

$$\begin{aligned}\Rightarrow \left(-\frac{3}{4}\right)^3 \times \left(\frac{2}{3}\right)^4\\[1em] = \frac{(-3) \times (-3) \times (-3)}{4 \times 4 \times 4} \times \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\\[1em] = -\frac{27}{64} \times \frac{16}{81}\\[1em] = -\frac{27 \times 16}{64 \times 81}\\[1em] = -\frac{1}{4 \times 3}\\[1em] = -\frac{1}{12}\end{aligned}$$

Hence, $\left(-\frac{3}{4}\right)^3 \times \left(\frac{2}{3}\right)^4 = -\frac{1}{12}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{3}{5}\right)^2 \times \left(-\frac{2}{3}\right)^3\\[1em] = \frac{3 \times 3}{5 \times 5} \times \frac{(-2) \times (-2) \times (-2)}{3 \times 3 \times 3}\\[1em] = \frac{9}{25} \times \left(-\frac{8}{27}\right)\\[1em] = -\frac{9 \times 8}{25 \times 27}\\[1em] = -\frac{8}{25 \times 3}\\[1em] = -\frac{8}{75}\end{aligned}$$

Hence, $\left(\frac{3}{5}\right)^2 \times \left(-\frac{2}{3}\right)^3 = -\frac{8}{75}$.

(i) Solving,

⇒ 2³ = 2 × 2 × 2 = 8

⇒ 3² = 3 × 3 = 9.

Since 9 > 8, therefore 3² > 2³.

Hence, the greater number is 3².

(ii) Solving,

⇒ 2⁵ = 2 × 2 × 2 × 2 × 2 = 32

⇒ 5² = 5 × 5 = 25.

Since 32 > 25, therefore 2⁵ > 5².

Hence, the greater number is 2⁵.

(iii) Solving,

⇒ 4³ = 4 × 4 × 4 = 64

⇒ 3⁴ = 3 × 3 × 3 × 3 = 81.

Since 81 > 64, therefore 3⁴ > 4³.

Hence, the greater number is 3⁴.

(iv) Solving,

⇒ 5⁴ = 5 × 5 × 5 × 5 = 625

⇒ 4⁵ = 4 × 4 × 4 × 4 × 4 = 1024.

Since 1024 > 625, therefore 4⁵ > 5⁴.

Hence, the greater number is 4⁵.

By prime factorisation of 512:

$$\begin{array}{l|r} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

⇒ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹.

Hence, 512 = 2⁹.

By prime factorisation of 1250:

$$\begin{array}{l|r} 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

⇒ 1250 = 2 × 5 × 5 × 5 × 5 = 2¹ × 5⁴.

Hence, 1250 = 2 × 5⁴.

By prime factorisation of 1458:

$$\begin{array}{l|r} 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

⇒ 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2¹ × 3⁶.

Hence, 1458 = 2 × 3⁶.

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.

By prime factorisation of 1350:

$$\begin{array}{l|r} 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

⇒ 1350 = 2 × 3 × 3 × 3 × 5 × 5 = 2¹ × 3³ × 5².

Hence, 1350 = 2 × 3³ × 5².

By prime factorisation of 1176:

$$\begin{array}{l|r} 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

⇒ 1176 = 2 × 2 × 2 × 3 × 7 × 7 = 2³ × 3¹ × 7².

Hence, 1176 = 2³ × 3 × 7².

(i) Solving,

⇒ (a + b)² = (2 + 3)² = 5² = 5 × 5 = 25.

Hence, (a + b)² = 25.

(ii) Solving,

⇒ (b - a)³ = (3 - 2)³ = 1³ = 1 × 1 × 1 = 1.

Hence, (b - a)³ = 1.

(iii) Solving,

⇒ (a × b)^a = (2 × 3)² = 6² = 6 × 6 = 36.

Hence, (a × b)^a = 36.

(iv) Solving,

⇒ (a × b)^b = (2 × 3)³ = 6³ = 6 × 6 × 6 = 216.

Hence, (a × b)^b = 216.

(i) By prime factorisation of 1024:

$$\begin{array}{l|r} 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

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

Hence, 1024 = 2¹⁰.

(ii) By prime factorisation of 343:

$$\begin{array}{l|r} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

⇒ 343 = 7 × 7 × 7 = 7³.

Hence, 343 = 7³.

(iii) By prime factorisation of 729:

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶.

Hence, 729 = 3⁶.

By prime factorisation of 27:

$$\begin{array}{l|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

⇒ 27 = 3 × 3 × 3 = 3³.

By prime factorisation of 32:

$$\begin{array}{l|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

⇒ 32 = 2 × 2 × 2 × 2 × 2 = 2⁵.

So,

⇒ 27 × 32 = 3^x × 2^y

⇒ 3³ × 2⁵ = 3^x × 2^y.

Comparing the powers of the same base on both sides, we get :

⇒ x = 3 and y = 5.

Hence, x = 3 and y = 5.

(i) Solving,

⇒ 64 × 625 = 2^a × 5^b

⇒ (2 × 2 × 2 × 2 × 2 × 2) × (5 × 5 × 5 × 5) = 2^a × 5^b

⇒ 2⁶ × 5⁴ = 2^a × 5^b.

Comparing the powers of the same base on both sides, we get :

⇒ a = 6 and b = 4.

Hence, a = 6 and b = 4.

(ii) Solving,

⇒ 2^b × 5^a = 2⁴ × 5⁶

⇒ (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5 × 5)

⇒ 16 × 15625

⇒ 250000.

Hence, 2^b × 5^a = 250000.

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

(i) In 5² = 25, base = 5 and index = 2

(ii) If index = 3x and base = 2y, the number = (2y)^3x

Solving,

2⁸ ÷ 2³

= $\frac{2^8}{2^3}$

= 2^{(8 - 3)}

= 2⁵

= 32

Hence, 2⁸ ÷ 2³ = 2⁵ = 32.

Solving,

$$\begin{aligned}\Rightarrow 2^3 ÷ 2^8 \\[1em] = \frac{2^3}{2^8} \\[1em] = \frac{1}{2^{8-3}} \\[1em] = \frac{1}{2^5} \\[1em] = \frac{1}{32}\end{aligned}$$

Hence, $2^3 ÷ 2^8 = \frac{1}{2^5} = \frac{1}{32}$.

Solving,

As any non-zero base raised to the power zero is equal to 1,

$$\begin{aligned}\Rightarrow (2^6)^0 \\[1em] = 1\end{aligned}$$

Hence, (2⁶)⁰ = 1.

Solving,

$$\begin{aligned}\Rightarrow (3^0)^6 \\[1em] = 1^6 \\[1em] = 1\end{aligned}$$

Hence, (3⁰)⁶ = 1.

Solving,

$$\begin{aligned}\Rightarrow 8^3 \times 8^{-5} \times 8^4 \\[1em] = 8^{3 + (-5) + 4} \\[1em] = 8^2 \\[1em] = 64\end{aligned}$$

Hence, 8³ × 8^-5 × 8⁴ = 8² = 64.

Solving,

$$\begin{aligned}\Rightarrow 5^4 \times 5^3 \div 5^5 \\[1em] = \frac{5^4 \times 5^3}{5^5} \\[1em] = \frac{5^7}{5^5} \\[1em] = 5^{7 - 5} \\[1em] = 5^2 \\[1em] = 25\end{aligned}$$

Hence, 5⁴ × 5³ ÷ 5⁵ = 5² = 25.

Solving,

$$\begin{aligned}\Rightarrow 5^4 \div 5^3 \times 5^5 \\[1em] = \frac{5^4 \times 5^5}{5^3} \\[1em] = \frac{5^9}{5^3} \\[1em] = 5^{9 - 3} \\[1em] = 5^6 \\[1em] = 15625\end{aligned}$$

Hence, 5⁴ ÷ 5³ × 5⁵ = 5⁶ = 15625.

Solving,

$$\begin{aligned}\Rightarrow 4^4 \div 4^3 \times 4^0 \\[1em] = \frac{4^4 \times 4^0}{4^3} \\[1em] = \frac{4^4}{4^3} \\[1em] = 4^{4 - 3} \\[1em] = 4^1 \\[1em] = 4\end{aligned}$$

Hence, 4⁴ ÷ 4³ × 4⁰ = 4.

Solving,

As any non-zero base raised to the power zero is equal to 1,

$$\begin{aligned}\Rightarrow (3^5 \times 4^7 \times 5^8)^0 \\[1em] = 1\end{aligned}$$

Hence, (3⁵ × 4⁷ × 5⁸)⁰ = 1.

Solving,

⇒ 2b⁶ · b³ · 5b⁴

⇒ (2 × 5) × b^{6 + 3 + 4}

⇒ 10b¹³.

Hence, 2b⁶ · b³ · 5b⁴ = 10b¹³.

Solving,

⇒ x²y³ · 6x⁵y · 9x³y⁴

⇒ (1 × 6 × 9) × x^{2 + 5 + 3} × y^{3 + 1 + 4}

⇒ 54x¹⁰y⁸.

Hence, x²y³ · 6x⁵y · 9x³y⁴ = 54x¹⁰y⁸.

Solving,

⇒ (-a⁵) (a²)

⇒ (-1) (a⁵) (a²)

⇒ -a^{5 + 2}

⇒ -a⁷.

Hence, (-a⁵) (a²) = -a⁷.

Solving,

⇒ (-y²) (-y³)

⇒ (-1) × (-1) × y^{2 + 3}

⇒ y⁵.

Hence, (-y²) (-y³) = y⁵.

Solving,

⇒ (-3)² (3)³

⇒ (-1)² (3)² (3)³

⇒ 3² × 3³

⇒ 3^{2 + 3}

⇒ 3⁵.

Hence, (-3)² (3)³ = 3⁵.

Solving,

⇒ (-4x) (-5x²)

⇒ (-4) × (-5) × x^{1 + 2}

⇒ 20x³.

Hence, (-4x) (-5x²) = 20x³.

Solving,

⇒ (5a²b) (2ab²) (a³b)

⇒ (5 × 2 × 1) × a^{2 + 1 + 3} × b^{1 + 2 + 1}

⇒ 10a⁶b⁴.

Hence, (5a²b) (2ab²) (a³b) = 10a⁶b⁴.

Solving,

⇒ x^{2a + 7} · x^{2a - 8}

⇒ x^{(2a + 7) + (2a - 8)}

⇒ x^{4a - 1}.

Hence, x^{2a + 7} · x^{2a - 8} = x^{4a - 1}.

Solving,

⇒ 3^y · 3² · 3^-4

⇒ 3^{y + 2 + (-4)}

⇒ 3^{y - 2}.

Hence, 3^y · 3² · 3^-4 = 3^{y - 2}.

Solving,

⇒ 2^4a · 2^3a · 2^-a

⇒ 2^{4a + 3a + (-a)}

⇒ 2^6a.

Hence, 2^4a · 2^3a · 2^-a = 2^6a.

Solving,

$$\begin{aligned}\Rightarrow 4x^2y^2 ÷ 9x^3y^3\\[1em] = \frac{4x^2y^2}{9x^3y^3}\\[1em] = \frac{4}{9} \times x^{2-3} \times y^{2-3}\\[1em] = \frac{4}{9} \times x^{-1} \times y^{-1}\\[1em] = \frac{4}{9xy}\end{aligned}$$

Hence, $4x^2y^2 ÷ 9x^3y^3 = \frac{4}{9xy}$.

Solving,

⇒ (10²)³ (x⁸)¹²

⇒ 10^{2 × 3} × x^{8 × 12}

⇒ 10⁶x⁹⁶.

Hence, (10²)³ (x⁸)¹² = 10⁶x⁹⁶.

Solving,

⇒ (a¹⁰)¹⁰ (1⁶)¹⁰

⇒ a^{10 × 10} × 1^{6 × 10}

⇒ a¹⁰⁰ × 1

⇒ a¹⁰⁰.

Hence, (a¹⁰)¹⁰ (1⁶)¹⁰ = a¹⁰⁰.

Solving,

⇒ (n²)² (-n²)³

⇒ n^{2 × 2} × (-1)³ × n^{2 × 3}

⇒ n⁴ × (-1) × n⁶

⇒ -n^{4 + 6}

⇒ -n¹⁰.

Hence, (n²)² (-n²)³ = -n¹⁰.

Solving,

⇒ - (3ab)² (-5a²bc⁴)²

⇒ - [3²a²b²] × [(-5)²a^{2 × 2}b²c^{4 × 2}]

⇒ - [9a²b²] × [25a⁴b²c⁸]

⇒ - (9 × 25) × a^{2 + 4} × b^{2 + 2} × c⁸

⇒ -225a⁶b⁴c⁸.

Hence, - (3ab)² (-5a²bc⁴)² = -225a⁶b⁴c⁸.

Solving,

As 0 multiplied with any number gives 0,

⇒ (-2)² × (0)³ × (3)³

⇒ 4 × 0 × 27

⇒ 0.

Hence, (-2)² × (0)³ × (3)³ = 0.

Solving,

⇒ (2a³)⁴ (4a²)²

⇒ [2⁴a^{3 × 4}] × [4²a^{2 × 2}]

⇒ [16a¹²] × [16a⁴]

⇒ (16 × 16) × a^{12 + 4}

⇒ 256a¹⁶.

Hence, (2a³)⁴ (4a²)² = 256a¹⁶.

Solving,

$$\begin{aligned}\Rightarrow (4x^2y^3)^3 ÷ (3x^2y^3)^3\\[1em] = \frac{(4x^2y^3)^3}{(3x^2y^3)^3}\\[1em] = \left(\frac{4x^2y^3}{3x^2y^3}\right)^3\\[1em] = \left(\frac{4}{3}\right)^3\\[1em] = \frac{64}{27}\end{aligned}$$

Hence, $(4x^2y^3)^3 ÷ (3x^2y^3)^3 = \frac{64}{27}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{1}{2x}\right)^3 \times (6x)^2\\[1em] = \frac{1}{2^3x^3} \times 6^2x^2\\[1em] = \frac{1}{8x^3} \times 36x^2\\[1em] = \frac{36x^2}{8x^3}\\[1em] = \frac{36}{8} \times x^{2-3}\\[1em] = \frac{9}{2} \times x^{-1}\\[1em] = \frac{9}{2x}\end{aligned}$$

Hence, $\left(\frac{1}{2x}\right)^3 \times (6x)^2 = \frac{9}{2x}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{1}{4ab^2c}\right)^2 ÷ \left(\frac{3}{2a^2bc^2}\right)^4\\[1em] = \frac{1}{4^2a^2b^4c^2} ÷ \frac{3^4}{2^4a^8b^4c^8}\\[1em] = \frac{1}{16a^2b^4c^2} ÷ \frac{81}{16a^8b^4c^8}\\[1em] = \frac{1}{16a^2b^4c^2} \times \frac{16a^8b^4c^8}{81}\\[1em] = \frac{16a^8b^4c^8}{16 \times 81 \times a^2b^4c^2}\\[1em] = \frac{1}{81} \times a^{8-2} \times b^{4-4} \times c^{8-2}\\[1em] = \frac{1}{81} \times a^6 \times b^0 \times c^6\\[1em] = \frac{a^6c^6}{81}\end{aligned}$$

Hence, $\left(\frac{1}{4ab^2c}\right)^2 ÷ \left(\frac{3}{2a^2bc^2}\right)^4 = \frac{a^6c^6}{81}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7}\\[1em] = \frac{5^3x^{21} \cdot 10^2x^4}{2^7x^{42}}\\[1em] = \frac{125x^{21} \cdot 100x^4}{128x^{42}}\\[1em] = \frac{125 \times 100}{128} \times \frac{x^{21 + 4}}{x^{42}}\\[1em] = \frac{12500}{128} \times x^{25 - 42}\\[1em] = \frac{3125}{32} \times x^{-17}\\[1em] = \frac{3125}{32x^{17}}\end{aligned}$$

Hence, $\frac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7} = \frac{3125}{32x^{17}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2}\\[1em] = \frac{7^2p^4q^{18}r^{10} \times 4^3p^3q^3r^3}{14^2p^{12}q^{20}r^8}\\[1em] = \frac{49p^4q^{18}r^{10} \times 64p^3q^3r^3}{196p^{12}q^{20}r^8}\\[1em] = \frac{49 \times 64}{196} \times \frac{p^{4 + 3}q^{18 + 3}r^{10 + 3}}{p^{12}q^{20}r^8}\\[1em] = \frac{3136}{196} \times p^{7 - 12} \times q^{21 - 20} \times r^{13 - 8}\\[1em] = 16 \times p^{-5} \times q^1 \times r^5\\[1em] = \frac{16qr^5}{p^5}\end{aligned}$$

Hence, $\frac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2} = \frac{16qr^5}{p^5}$.

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow \frac{(2^3)^5 \times 5^4}{4^3 \times 5^2}\\[1em] = \frac{2^{3 \times 5} \times 5^4}{(2^2)^3 \times 5^2}\\[1em] = \frac{2^{15} \times 5^4}{2^6 \times 5^2}\\[1em] = 2^{15-6} \times 5^{4-2}\\[1em] = 2^9 \times 5^2\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5}\\[1em] = \frac{(2^2 \times 3^2) \times (2^2 \times 3^2) \times 3^6}{(2^2 \times 3)^3 \times 3^5}\\[1em] = \frac{2^4 \times 3^{10}}{2^6 \times 3^3 \times 3^5}\\[1em] = \frac{2^4 \times 3^{10}}{2^6 \times 3^8}\\[1em] = 2^{4-6} \times 3^{10-8}\\[1em] = 2^{-2} \times 3^2\\[1em] = \frac{3^2}{2^2}\\[1em] = \left(\frac{3}{2}\right)^2\end{aligned}$$

Hence, $\frac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5} = \left(\frac{3}{2}\right)^2$.

Solving,

$$\begin{aligned}\Rightarrow -\frac{128}{2187}\\[1em] = -\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}\\[1em] = -\frac{2^7}{3^7}\\[1em] = -\left(\frac{2}{3}\right)^7\\[1em] = \left(-\frac{2}{3}\right)^7\end{aligned}$$

Hence, $-\frac{128}{2187} = \left(-\frac{2}{3}\right)^7$.

Solving,

$$\begin{aligned}\Rightarrow \frac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8}\\[1em] = a^{-7-3} \times b^{-7-(-5)} \times c^{5-(-3)} \times d^{4-8}\\[1em] = a^{-10} \times b^{-2} \times c^8 \times d^{-4}\\[1em] = \frac{c^8}{a^{10} \times b^2 \times d^4}\end{aligned}$$

Hence, $\frac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8} = \frac{c^8}{a^{10}b^2d^4}$.

Solving,

$$\begin{aligned}\Rightarrow (a^3b^{-5})^{-2} \\[1em] = a^{3 \times (-2)} \times b^{-5 \times (-2)} \\[1em] = a^{-6} \times b^{10} \\[1em] = \frac{b^{10}}{a^6}\end{aligned}$$

Hence, $(a^3b^{-5})^{-2} = \frac{b^{10}}{a^6}$.

Solving,

$$\begin{aligned}\Rightarrow 6^{-2} ÷ (4^{-2} \times 3^{-2})\\[1em] = \frac{1}{6^2} ÷ \left(\frac{1}{4^2} \times \frac{1}{3^2}\right)\\[1em] = \frac{1}{36} ÷ \left(\frac{1}{16} \times \frac{1}{9}\right)\\[1em] = \frac{1}{36} ÷ \frac{1}{144}\\[1em] = \frac{1}{36} \times 144\\[1em] = 4\end{aligned}$$

Hence, 6^-2 ÷ (4^-2 × 3^-2) = 4.

Solving,

$$\begin{aligned}\Rightarrow \left[\left(\frac{5}{6}\right)^2 \times \frac{9}{4}\right] \div \left[\left(-\frac{3}{2}\right)^2 \times \frac{125}{216}\right]\\[1em] = \left[\frac{25}{36} \times \frac{9}{4}\right] \div \left[\frac{9}{4} \times \frac{125}{216}\right]\\[1em] = \frac{25}{16} \div \frac{125}{96}\\[1em] = \frac{25}{16} \times \frac{96}{125}\\[1em] = \frac{25 \times 96}{16 \times 125}\\[1em] = \frac{6}{5}\\[1em] = 1\frac{1}{5}\end{aligned}$$

Hence, the value is $1\frac{1}{5}$.

Solving,

$$\begin{aligned}\Rightarrow 5^3 \times 3^2 + (17)^0 \times 7^3 \\[1em] = (125 \times 9) + (1 \times 343) \\[1em] = 1125 + 343 \\[1em] = 1468\end{aligned}$$

Hence, 5³ × 3² + (17)⁰ × 7³ = 1468.

Solving,

$$\begin{aligned}\Rightarrow 2^5 \times 15^0 + (-3)^3 - \left(\frac{2}{7}\right)^{-2}\\[1em] = (32 \times 1) + (-27) - \left(\frac{7}{2}\right)^2\\[1em] = 32 - 27 - \frac{49}{4}\\[1em] = 5 - \frac{49}{4}\\[1em] = \frac{20}{4} - \frac{49}{4}\\[1em] = \frac{20 - 49}{4}\\[1em] = -\frac{29}{4}\end{aligned}$$

Hence, the value is $-\frac{29}{4}$.

Solving,

$$\begin{aligned}\Rightarrow (2^2)^0 + 2^{-4} \div 2^{-6} + \left(\frac{1}{2}\right)^{-3} \\[1em] = 1 + 2^{-4 - (-6)} + 2^3 \\[1em] = 1 + 2^2 + 8 \\[1em] = 1 + 4 + 8 \\[1em] = 13\end{aligned}$$

Hence, the value is 13.

Solving,

$$\begin{aligned}\Rightarrow 5^n \times 25^{n-1} ÷ (5^{n-1} \times 25^{n-1})\\[1em] = \frac{5^n \times 25^{n-1}}{5^{n-1} \times 25^{n-1}}\\[1em] = \frac{5^n}{5^{n-1}}\\[1em] = 5^{n-(n-1)}\\[1em] = 5^1\\[1em] = 5\end{aligned}$$

Hence, 5ⁿ × 25^n-1 ÷ (5^n-1 × 25^n-1) = 5.

Solving,

$$\begin{aligned}\Rightarrow m^2 + n^2 - 2mn \\[1em] = (-2)^2 + (2)^2 - 2 \times (-2) \times (2) \\[1em] = 4 + 4 - (-8) \\[1em] = 8 + 8 \\[1em] = 16\end{aligned}$$

Hence, m² + n² - 2mn = 16.

Solving,

$$\begin{aligned}\Rightarrow m^n + n^m\\[1em] = (-2)^2 + (2)^{-2}\\[1em] = 4 + \frac{1}{2^2}\\[1em] = 4 + \frac{1}{4}\\[1em] = \frac{16}{4} + \frac{1}{4}\\[1em] = \frac{17}{4}\\[1em] = 4\frac{1}{4}\end{aligned}$$

Hence, $m^n + n^m = 4\frac{1}{4}$.

Solving,

$$\begin{aligned}\Rightarrow 6m^{-3} + 4n^2\\[1em] = 6 \times (-2)^{-3} + 4 \times (2)^2\\[1em] = \frac{6}{(-2)^3} + 4 \times 4\\[1em] = \frac{6}{-8} + 16\\[1em] = -\frac{3}{4} + 16\\[1em] = -\frac{3}{4} + \frac{64}{4}\\[1em] = \frac{61}{4}\\[1em] = 15\frac{1}{4}\end{aligned}$$

Hence, $6m^{-3} + 4n^2 = 15\frac{1}{4}$.

Solving,

$$\begin{aligned}\Rightarrow 2n^3 - 3m \\[1em] = 2 \times (2)^3 - 3 \times (-2) \\[1em] = 2 \times 8 - (-6) \\[1em] = 16 + 6 \\[1em] = 22\end{aligned}$$

Hence, 2n³ - 3m = 22.

(i) False. 8 × 8¹⁵ = 8^{1 + 15} = 8¹⁶, whereas 64¹⁶ = (8²)¹⁶ = 8³². Since 8¹⁶ ≠ 8³², the statement is false.

(ii) False. 16⁸ ÷ 4² = (4²)⁸ ÷ 4² = 4¹⁶ ÷ 4² = 4¹⁴, which is not equal to 4⁶.

(iii) True. Any non-zero base raised to the power zero is equal to 1, so 27⁰ = 1 = 54903⁰.

(iv) True. When n is an even whole number, (-1)ⁿ = 1.

(v) False. Here the expression is read as (-1)ⁿ. For an even whole number n, (-1)ⁿ = 1 and not -1; it equals -1 only when n is odd. So the value is not -1 for every odd or even whole number n.

(vi) False. $(-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{-27} = -\frac{1}{27}$, which is not +9.

(vii) False. $4^{-4} = \frac{1}{4^4} = \frac{1}{256}$, which is not -16.

Solving,

$$\begin{aligned}\Rightarrow \left(-\frac{2}{3}\right)^{-3}\\[1em] = \left(-\frac{3}{2}\right)^{3}\\[1em] = \frac{(-3) \times (-3) \times (-3)}{2 \times 2 \times 2}\\[1em] = \frac{-27}{8}\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (-3)^2 ÷ \left(-\frac{1}{2}\right)^3\\[1em] = 9 ÷ \left(-\frac{1}{8}\right)\\[1em] = 9 \times (-8)\\[1em] = -72\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

⇒ (-2)⁵ = (-2) × (-2) × (-2) × (-2) × (-2) = -32.

Reciprocal of -32 = $\frac{1}{-32} = -\frac{1}{32}$

Hence, Option 2 is the correct option.

Solving,

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

Comparing the powers of the same base,

⇒ -(x + 6) = -2

⇒ x + 6 = 2

⇒ x = -4

Hence, Option 4 is the correct option.

As any non-zero base raised to the power zero is equal to 1,

⇒ 4⁰ + 6⁰ - 8⁰

⇒ 1 + 1 - 1

⇒ 1.

Hence, Option 2 is the correct option.

Solving,

⇒ (2³)² ÷ (2²)⁴

= 2^{3 × 2} ÷ 2^{2 × 4}

= 2⁶ ÷ 2⁸

= 2^6-8

= 2^-2

= $\frac{1}{2^2} = \frac{1}{4}$

Hence, Option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \left(-\frac{2}{3}\right)^x = -\frac{243}{32}\\[1em] \Rightarrow \left(-\frac{2}{3}\right)^x = -\frac{3^5}{2^5}\\[1em] \Rightarrow \left(-\frac{2}{3}\right)^x = \left(-\frac{3}{2}\right)^5\\[1em] \Rightarrow \left(-\frac{2}{3}\right)^x = \left(-\frac{2}{3}\right)^{-5}\end{aligned}$$

Comparing the powers of the same base,

⇒ x = -5.

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (-4)^3 ÷ (4)^4\\[1em] = \frac{(-4)^3}{4^4}\\[1em] = \frac{-64}{256}\\[1em] = -\frac{1}{4}\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

⇒ 7⁰ × 5 - (-2)³ - 8⁰

⇒ (1 × 5) - (-8) - 1

⇒ 5 + 8 - 1

⇒ 12.

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{9}{10}\right)^2 \times \left(\frac{10}{9}\right)^5 = \left(\frac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\frac{9}{10}\right)^2 \times \left(\frac{9}{10}\right)^{-5} = \left(\frac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\frac{9}{10}\right)^{2 - 5} = \left(\frac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\frac{9}{10}\right)^{-3} = \left(\frac{9}{10}\right)^{1-m}\end{aligned}$$

Comparing the powers of the same base,

⇒ 1 - m = -3

⇒ m = 4

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{3}{5}\right)^{-1} \div \left(\frac{-5}{2}\right)^{-1}\\[1em] = \frac{5}{3} \div \left(-\frac{2}{5}\right)\\[1em] = \frac{5}{3} \times \left(-\frac{5}{2}\right)\\[1em] = -\frac{25}{6}\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

⇒ (-3)^{x - 1} = -243

⇒ (-3)^{x - 1} = (-3)⁵

Comparing the powers of the same base,

⇒ x - 1 = 5

⇒ x = 6

Now,

⇒ (243)^x-6 = (243)^{6 - 6} = (243)⁰ = 1.

Hence, Option 4 is the correct option.

⇒ {(5²)³ × 5⁴} ÷ 5³

⇒ {5^{2 × 3} × 5⁴} ÷ 5³

⇒ {5⁶ × 5⁴} ÷ 5³

⇒ 5^{6 + 4} ÷ 5³

⇒ 5¹⁰ ÷ 5³

⇒ 5^{10 - 3}

⇒ 5⁷.

Hence, Option 2 is the correct option.

Solving,

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

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (-9)^{-1} ÷ x = (18)^{-1}\\[1em] \Rightarrow -\frac{1}{9} ÷ x = \frac{1}{18}\\[1em] \Rightarrow -\frac{1}{9} \times \frac{1}{x} = \frac{1}{18}\\[1em] \Rightarrow \frac{1}{x} = \frac{1}{18} \times (-9)\\[1em] \Rightarrow \frac{1}{x} = -\frac{9}{18}\\[1em] \Rightarrow \frac{1}{x} = -\frac{1}{2}\\[1em] \Rightarrow x = -2\end{aligned}$$

Hence, Option 2 is the correct option.

By prime factorisation,

$$\begin{array}{l|r} 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹¹, not 2¹².

Thus, Statement 1 is false.

For any number 'a' and a positive integer 'n', aⁿ = a × a × a × .... a (n times) is the correct definition.

Thus, Statement 2 is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, Option 4 is the correct option.

Solving,

$$\begin{aligned}\left(a^2\right)^{\frac{1}{4}} \div \left(a^3\right)^{\frac{1}{6}}\\[1em] = a^{2 \times \frac{1}{4}} \div a^{3 \times \frac{1}{6}}\\[1em] = a^{\frac{1}{2}} \div a^{\frac{1}{2}}\\[1em] = a^{\frac{1}{2} - \frac{1}{2}}\\[1em] = a^0\\[1em] = 1\end{aligned}$$

Thus, Statement 1 is true.

The power law states that (aⁿ)^m = a^mn, which is correct.

Thus, Statement 2 is true.

∴ Both the statements are true.

Hence, Option 1 is the correct option.

Solving,

⇒ (9⁰ + 8⁰) ÷ (72)⁰

⇒ (1 + 1) ÷ 1

⇒ 2 ÷ 1

⇒ 2.

Since the value is 2 and not 1,

Thus, Assertion (A) is false.

Any non-zero base raised to the power zero is equal to 1, which is a correct statement.

Thus, Reason (R) is true.

∴ A is false, R is true.

Hence, Option 2 is the correct option.

Here -1ⁿ means -(1ⁿ). Since 1ⁿ = 1 for every whole number n, -1ⁿ = -1 for all whole numbers n.

Thus, Assertion (A) is true.

(-1)ⁿ is +1 when n is even and -1 when n is odd. So (-1)ⁿ is not equal to -1ⁿ for all n ∈ W.

Thus, Reason (R) is false.

∴ A is true, R is false.

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\left(-\frac{3}{4}\right) \times \left(-\frac{3}{4}\right) \times \left(-\frac{3}{4}\right) \times \dots \text{ up to 10 terms}\\[1em] = \left(-\frac{3}{4}\right)^{10}\\[1em] = \frac{(-3)^{10}}{4^{10}}\\[1em] = \frac{3^{10}}{(2^2)^{10}}\\[1em] = \frac{3^{10}}{2^{20}}\end{aligned}$$

Comparing with $\frac{3^a}{2^b}$, we get a = 10 and b = 20.

⇒ a - b = 10 - 20 = -10.

Thus, Assertion (A) is true.

For an expression in the form of a fraction with the same power, $\left(\frac{p}{q}\right)^n = \frac{p^n}{q^n}$, which is correct.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.