It is a prerequisite here. There, we stated the 7 laws of indices and the two Rules used for solving problems.

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

We provided a few Solved Examples and Problems for Practice with Answers to help to apply the 7 Laws and the 2 Rules in solving problems.

Here, we provide many more Solved Examples and Exercises with answers.

Studying the worked out problems will help remember and apply the 7 Laws of exponents and the 2 Rules.

Practice makes one perfect. This is especially true for remembering Algebra Formulas (Math Formulas). So, take the exercises seriously and practice solving the problems.

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Set of Solved Examples of Math Exponents : Application of Laws of Exponents

Solved Example 1 of Math Exponents

Evaluate (-1⁄2)^{5}⁄(-1⁄2)^{4} + (-1⁄8)⁄(-1⁄4)

Solution to Example 1 of Math Exponents:

We know a^{m}⁄a^{n} = a^{m - n} Applying this Law, we get (-1⁄2)^{5}⁄(-1⁄2)^{4} + (-1⁄8)⁄(-1⁄4) = (-1⁄2)^{5 - 4} + (-1⁄8) x (4⁄-1) = (-1⁄2)^{1} + 1⁄2 = -1⁄2 + 1⁄2 = 0. Ans.

Solved Example 2 of Math Exponents

What is (-1)^{401}?

Solution to Example 2 of Math Exponents:

Let A = (-1)^{401} A = (-1)^{401} = (-1) ^{ 400 + 1} Interchanging L.H.S. and R.H.S. in Law 1, we have a^{m + n} = a^{m} x a^{n} Applying this here, we get A = (-1)^{400} x (-1)^{1} = (-1)^{2 x 200} x (-1)^{1} Interchanging L.H.S. and R.H.S. in Law 2, we have a^{mn} = (a^{m})^{n} Applying this here, we get A = {(-1)^{2}}^{200} x (-1)^{1} We know (-1)^{2} = (-1) x (-1) = 1. and (-1)^{1} = -1. ∴ A = 1^{200} x (-1 ) We know 1power any number is 1. ∴ A = 1 x (-1 ) = -1. Thus (-1)^{401} = -1. Ans.

You might have noted (-1)^{odd number} = -1 and (-1)^{even number} = 1.

Solved Example 3 of Math Exponents

Given that 5^{x} = 1000, find 5^{x + 2} and 5^{x - 2}

Solution to Example 3 of Math Exponents:

Let A = 5^{x + 2} and B = 5^{x - 2} Interchanging L.H.S. and R.H.S. in Law 1, we have a^{m + n} = a^{m} x a^{n} Applying this here, we get A = 5^{x + 2} = 5^{x} x 5^{2} By data 5^{x} = 1000. Using this here, we get A = 1000 x 5^{2} = 1000 x 25 = 25000. Thus 5^{x + 2} = 25000. Ans.

Interchanging L.H.S. and R.H.S. in Law 4, we have a^{m - n} = a^{m}⁄a^{n} Applying this here, we get B = 5^{x - 2} = 5^{x}⁄5^{2} By data 5^{x} = 1000. Using this here, we get B = 1000⁄5^{2} = 1000⁄25 = 40. Thus 5^{x - 2} = 40. Ans.

Here the exponents are 3 on the L.H.S. and 3^{3} on the R.H.S. To find x which is in the base of L.H.S., let us make the exponents same. We know 3^{3} = 3 x 3 x 3 = 3 x 9 = 9 x 3. R.H.S. = 4^{33} = 4^{9 x 3} Interchanging L.H.S. and R.H.S. in Law 2, we have a^{mn} = (a^{m})^{n} Applying this here, we get R.H.S. = 4^{9 x 3} = (4^{9})^{3} L.H.S. = (4x)^{3} So, we have (4x)^{3} = (4^{9})^{3} Since the exponents are equal, the bases should be equal. ∴ (4x) = (4^{9}) ⇒ x = 4^{9}⁄4 = (4^{9})⁄4^{1} We know a^{m}⁄a^{n} = a^{m - n} Applying this here, we get x = 4^{9 - 1} = 4^{8}. Ans.

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If p⁄q = (2⁄3)^{3} + (3⁄2)^{-3}, what is (p⁄q)^{-10} ?

Solution to Example 5 of Math Exponents:

We know (a⁄b)^{-n} = (b⁄a)^{n}( See Law 3) Applying this, (3⁄2)^{-3} = (2⁄3)^{3}. By data, p⁄q = (2⁄3)^{3} + (3⁄2)^{-3}. = (2⁄3)^{3} + (2⁄3)^{3} = 2{(2⁄3)^{3}} We know (a⁄b)^{m} = a^{m}⁄b^{m} (See Law 7) Applying this, (2⁄3)^{3} = (2^{3})⁄(3^{3}) = 8⁄27 ∴ p⁄q = 2{(2⁄3)^{3}} = 2{8⁄27} = 16⁄27 ⇒ (p⁄q)^{-10} = (16⁄27)^{-10} We know (a⁄b)^{-n} = (b⁄a)^{n}.( See Law 3) Applying this, (16⁄27)^{-10} = (27⁄16)^{10}. ∴ (p⁄q)^{-10} = (16⁄27)^{-10} = (27⁄16)^{10}. Ans.

Exercise of Math Exponents : Application of Laws of Exponents

Problems on Math Exponents :

Simplify (i) (4x^{3})^{2} x (3x^{3})^{2} x (3xy)^{4} (ii) a^{x + y - z} x a^{y + z - x} x a^{z + x - y}

With what should you multiply 2^{a} to get 2^{a + 2} and 2^{a - 2}

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

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

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