Law of Indices (घाताङ्कको नियमहरू)
100% Unomitted Solutions: Textbook Exercises Q1 to Q7 & Past BLE Exam Questions
1 Simplify using the laws of indices:
$$= 3^{4 + 3} = 3^7 = 2187$$
$$= x^{3 + 5} = x^8$$
$$= a \times (b^4 \times b^3) = a \times b^{4+3} = ab^7$$
$$= (a^2 \times a) \times (b \times b^3) = a^{2+1} b^{1+3} = a^3b^4$$
$$= (3 \times 2) \times (x^4 \times x^3) = 6 \times x^{4+3} = 6x^7$$
$$= (-2 \times 3) \times (x^4 \times x^3) = -6x^{4+3} = -6x^7$$
$$= (a \times a^3 \times a^2) \times (b \times b^3 \times b) = a^{1+3+2} b^{1+3+1} = a^6b^5$$
2 Simplify using quotient rule:
$$= 4^{4-2} = 4^2 = 16$$
$$= x^{8-5} = x^3$$
$$= (a^4 \div a^3) \times (b^4 \div b^3) = a^{4-3} b^{4-3} = ab$$
$$= (x^6 \div x^3) \times (y^3 \div y^3) = x^{6-3} y^{3-3} = x^3 y^0 = x^3$$
$$= (8 \div 2) \times (x^4 \div x^3) = 4x^{4-3} = 4x$$
$$= (16 \div 8) \times (x^4 \div x^3) = 2x^{4-3} = 2x$$
3 Simplify using power & zero exponent rules:
$$= 1 \quad [\text{Since } x^0 = 1]$$
$$= 2^3 \times b^3 = 8b^3$$
$$= (-3)^4 \times x^4 = 81x^4$$
$$= (-4)^3 \times a^3 \times (b^2)^3 = -64a^3b^6$$
$$= 3^2 \times (a^3)^2 \times (b^2)^2 = 9a^6b^4$$
$$= \frac{(x^2)^2}{(y^2)^2} = \frac{x^4}{y^4}$$
$$= (3xy)^{2-1} = 3xy$$
$$= \frac{a^{4n-2}}{a^{4n-2}} = a^{(4n-2)-(4n-2)} = a^0 = 1$$
4 Simplify numeric expressions:
$$= \frac{2^2 \times (2^2)^2}{(2^3)^2} = \frac{2^2 \times 2^4}{2^6} = \frac{2^6}{2^6} = 1$$
$$= \frac{5^3 \times (5^3)^3}{(5^2)^3} = \frac{5^3 \times 5^9}{5^6} = 5^{12-6} = 5^6 = 15625$$
$$= \frac{(2^2)^4 \times 5^5}{(5^2)^3 \times (2^4)^2} = \frac{2^8 \times 5^5}{5^6 \times 2^8} = 5^{5-6} = \frac{1}{5}$$
5 Fill in the boxes ($\square$):
$$(2^2)^\square = (2^3)^2 \implies 2^{2\square} = 2^6$$
$$2\square = 6 \implies \square = 3$$
$$9x^{2\square} = 9x^6 \implies 2\square = 6$$
$$\implies \square = 3$$
$$\text{Since } (4x)^0 = 1 \implies \square = 0$$
6 Prove the following statements:
$$\text{LHS} = \frac{x^{2m+2n+4}}{x^{2m+2n+2}} = x^{(2m+2n+4)-(2m+2n+2)} = x^2 = \text{RHS (Proved)}$$
$$\text{LHS} = \frac{x^{(p-q+1+q-r+1+r-p+1)}}{x^3} = \frac{x^3}{x^3} = 1 = \text{RHS (Proved)}$$
$$\text{LHS} = x^{(a-b)(a+b)} \times x^{(b-c)(b+c)} \times x^{(c-a)(c+a)} = x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}$$
$$= x^{a^2-b^2+b^2-c^2+c^2-a^2} = x^0 = 1 = \text{RHS (Proved)}$$
7 If $a = 2, b = 3, c = 1, m = 4$ and $n = 5$, evaluate:
$$\text{Numerator} = 2^4 \times 3^5 \times 1^6 = 16 \times 243 \times 1 = 3888$$
$$\text{Denominator} = 4^2 \times 5^3 \times (3 \times 2)^1 = 16 \times 125 \times 6 = 12000$$
$$\text{Value} = \frac{3888}{12000} = \frac{81}{250}$$
$$a + b + c = 2 + 3 + 1 = 6, \quad m + n = 4 + 5 = 9$$
$$\text{Value} = 6^9 \div 9^6 = \frac{(2 \times 3)^9}{(3^2)^6} = \frac{2^9 \times 3^9}{3^{12}} = \frac{2^9}{3^3} = \frac{512}{27}$$
