There, we stated the 7 laws of indices and the two Rules used for solving problems.

We also gave Links to the explanations and proofs of the 7 laws.

Here, in Addition of Exponents, we apply the two Laws of Exponents where product/quotient of powers of the same base = addition/difference of exponents to that base. Solved Examples, Exercise, provided, help in mastering the Laws.

Solved Example 1 : Addition of Exponents

Simplify: x^{2} x (-4x)^{4} x y^{3} x 2y^{-2}

Solution to Example 1 of Addition of Exponents :

Let A = x^{2} x (-4x)^{4} x y^{3} x 2y^{-2} We know (-4x)^{4} = (-4)^{4} x x^{4}. Grouping constant terms and variables of one type at one place, we get A = {(-4)^{4} x 2} x (x^{2} x x^{4}) x (y^{3} x y^{-2}) We know even power of minus will become plus. So (-4)^{4} = 4^{4} = 4 x 4 x 4 x 4 = 256. We know a^{m} x a^{n} = a^{m + n} Applying this Law, we get A = (256 x 2) x (x^{2 + 4}) x {y^{3 + (-2)}} = 512 x x^{6} x y^{1} = 512x^{6}y Thus x^{2} x (-4x)^{4} x y^{3} x 2y^{-2} = 512x^{6}y. Ans.

Solved Example 2 : Addition of Exponents

Simplify: (2x^{2}y) x (-3x^{2}y^{2})^{3} x (xy)^{-1}

Solution to Example 2 of Addition of Exponents :

Let A = (2x^{2}y) x (-3x^{2}y^{2})^{3} x (xy)^{-1} We know (ab)^{m} = a^{m} x b^{m} (See Law 6) Applying this Law, we get A = (2x^{2}y) x (-3)^{3}(x^{2})^{3}(y^{2})^{3} x (x)^{-1}(y)^{-1} We know (a^{m})^{n} = a^{mn} (See Law 2) Applying this Law, we get A = (2x^{2}y) x (-3) x (-3) x (-3) (x^{2 x 3})(y^{2 x 3}) x (x)^{-1}(y)^{-1} Grouping constant terms and variables of one type at one place, we get A = (2) x (-3) x (-3) x (-3) (x^{2})(x^{2 x 3})(x^{-1})(y )(y^{2 x 3})(y^{-1}) ⇒ A = (2) x (-27)(x^{2})(x^{6})(x^{-1})(y )(y^{6})(y^{-1}) We know a^{m} x a^{n} = a^{m + n} Applying this Law, we get A = (-54){x^{2 + 6 + (-1)}}{y^{1 + 6 + (-1)}} ⇒ A = (-54){x^{7}}{y^{6}} Thus 2x^{2}y) x (-3x^{2}y^{2})^{3} x (xy)^{-1} = -54x^{7}y^{6}. Ans.

Solved Example 3 : Addition of Exponents

Simplify: (6a^{9}b^{4}c^{2})⁄(2a^{7}b^{2}c)

Solution to Example 3 of Addition of Exponents :

Let A = (6a^{9}b^{4}c^{2})⁄(2a^{7}b^{2}c) Grouping constant terms and variables of one type at one place, we get A = (6⁄2)(a^{9}⁄a^{7})(b^{4}⁄b^{2})(c^{2}⁄c) We know a^{m}⁄a^{n} = a^{m - n} Applying this here, we get A = (3)(a^{9 - 7})(b^{4 - 2})(c^{2 - 1}) = 3a^{2}b^{2}c^{1} Thus (6a^{9}b^{4}c^{2})⁄(2a^{7}b^{2}c) = 3a^{2}b^{2}c. Ans.

Let A = (3a^{2}b^{3}c^{4})^{2}⁄(2a^{4}b^{3}c^{2})^{2} We know (ab)^{m} = a^{m} x b^{m} (See Law 6) Applying this Law, we get A = {3^{2}(a^{2})^{2}(b^{3})^{2}(c^{4})^{2}}⁄{2^{2}(a^{4})^{2}(b^{3})^{2}(c^{2})^{2}} We know (a^{m})^{n} = a^{mn} (See Law 2) Applying this Law, we get A = {9(a^{2 x 2})(b^{3 x 2})(c^{4 x 2})}⁄{4(a^{4 x 2})(b^{3 x 2})(c^{2 x 2})} ⇒ A = (9a^{4}b^{6}c^{8})⁄(4a^{8}b^{6}c^{4}) Grouping constant terms and variables of one type at one place, we get A = (9⁄4)(a^{4}⁄a^{8})(b^{6}⁄b^{6})(c^{8}⁄c^{4}) We know a^{m}⁄a^{n} = a^{m - n} Applying this here, we get A = (9⁄4)(a^{4 - 8})(b^{6 - 6})(c^{8 - 4}) ⇒ A = (9⁄4)(a^{-4})(b^{0})(c^{4}) We know a^{-4} = 1⁄a^{4} and b^{0} = 1. ∴ A = (9⁄4)(1⁄a^{4})(1)(c^{4}) ⇒ A = (9c^{4})⁄(4a^{4}) Thus (3a^{2}b^{3}c^{4})^{2}⁄(2a^{4}b^{3}c^{2})^{2} = (9c^{4})⁄(4a^{4}). Ans.

Solved Example 5 : Addition of Exponents

Simplify: Show that a^{x(y - z)} x a^{y(z - x)} x a^{z(x - y)} = 1.

Solution to Example 5 of Addition of Exponents :

We know a^{m} x a^{n} = a^{m + n} and we can apply this Law for 3 terms also. ∴ L.H.S. = a^{x(y - z)} x a^{y(z - x)} x a^{z(x - y)} = a^{x(y - z) + y(z - x) + z(x - y)} = a^{(xy - xz + yz - yx + zx - zy)} Since xy = yx, yz = zy, zx = xz, we have L.H.S = a^{(xy - xy + yz - yz + zx - zx)} = a^{(0 + 0 + 0)} = a^{(0)} = a^{0}. We know a^{0} = 1. By applying this Law, we get L.H.S. = 1 = R.H.S. (proved)

Solved Example 6 : Addition of Exponents

Simplify: Simplify (x^{a}⁄x^{b})^{c} x (x^{b}⁄x^{c})^{a} x (x^{c}⁄x^{a})^{b}

Solution to Example 6 of Addition of Exponents :

Let A = (x^{a}⁄x^{b})^{c} x (x^{b}⁄x^{c})^{a} x (x^{c}⁄x^{a})^{b} We know a^{m}⁄a^{n} = a^{m - n} Applying this Law, we get A = (x^{a - b})^{c} x (x^{b - c})^{a} x (x^{c - a})^{b} We know (a^{m})^{n} = a^{mn} Applying this Law, we get A = x^{(a - b)c} x x^{(b - c)a} x x^{(c - a)b} We know a^{m} x a^{n} = a^{m + n} Applying this Law, we get A = x^{(a - b)c + (b - c)a + (c - a)b} ⇒ A = x^{(ac - bc) + (ba - ca) + (cb - ab)} regrouping, we get A = x^{(ac - ca) + (ba - ab ) + (cb - bc)} Since ac = ca, ba = ab, cb = bc, we get A = x^{(0) + (0) + (0)} = x^{0} We know a^{0} = 1. Applying this, we get A = 1. Thus (x^{a}⁄x^{b})^{c} x (x^{b}⁄x^{c})^{a} x (x^{c}⁄x^{a})^{b} = 1. Ans.

For More Solved Examples on application of Laws of Exponents, go to Math Exponents.

Get The Best Grades With the Least Amount of Effort : Addition of Exponents

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Simplify: (i) 5^{4} x 5^{3} x 5^{-5} (ii) 3^{0} x 4^{3} x 2^{-5} (iii) (-2)^{4} x (-4)^{2} x 2^{-2} x 4^{2} (iv) 3^{6} x 6^{2} x 2^{-4}

Simplify: (i) 3^{8}⁄3^{5} (ii) (2^{5} x 2^{4})⁄2^{13} (iii) 1⁄5^{-3} (iv) (-3)^{-4}⁄(-3)^{-8}

Simplify: (i) {(3^{2})^{4}}^{6} (ii) {(-5)^{4} x 7^{5}}^{3} (iii) {(-8)^{-4} x 2^{-8}}^{2} (iv) {(4)^{-2} x 6^{-2}}^{2} x 3^{-4}

Evaluate : (i) (3^{-3} x 5^{-2} x 3^{6})⁄(25 x 3^{-2} x 5^{-5}) (ii) {(2^{3})^{2} x (-5)^{2}}⁄{(5^{2})^{2} x 16} (iii) {(2^{-3})^{2} x (3^{2})^{-3} x (5^{-4})^{2} }⁄{(2^{-2})^{5} x (3^{3})^{-2} x (5^{-3})^{4} } (iv) (l^{-4} x m^{3} x n^{-2} x z^{6})⁄(l^{-1} x m^{-2} x n^{3} x z^{0})

For the Answers, see at the bottom of the page.

For more problems for practice on Application of Laws of exponents, go to Math Exponents.

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