There, we provided the explanation for Rational Exponents.

We discussed how we can apply the same 7 Laws and the 2 Rules given for whole number Exponents can be applied for Fractional Exponents.

Here, we apply the Laws to simplify a given expression or to prove a given relationship, by applying Laws of Exponents.

Solved Example 1 of Exponent Rules

Simplify (3^{n} x 9^{n + 1})⁄(3^{n - 1} x 9^{n - 1})

Solution to Example 1 of Exponent Rules:

Let A = (3^{n} x 9^{n + 1})⁄(3^{n - 1} x 9^{n - 1}) Writing powers with same base at one place, we get A = {(3^{n})⁄(3^{n - 1})} x {(9^{n + 1})⁄(9^{n - 1})} We know a^{m}⁄a^{n} = a^{m - n} Applying this here, we get A = {3^{n - (n - 1)}} x {9^{(n + 1) - (n - 1)}} = {3^{n - n + 1)}} x {9^{(n + 1 - n + 1)}} = (3^{1}) x (9^{2}) = 3 x 9 x 9 = 243. Ans.

Solved Example 2 of Exponent Rules

Simplify { 3 x (27)^{n+1} + 9 x (3)^{3n-1}}⁄{ 8 x 3^{3n} - 5 x (27)^{n}}

Solution to Example 2 :

Let A = { 3 x (27)^{n+1} + 9 x (3)^{3n-1}}⁄{ 8 x 3^{3n} - 5 x (27)^{n}} We know a^{m + n} = a^{m} x a^{n} Applying this here, we get A = { 3 x (27)^{n} x (27)^{1} + 9 x (3)^{3n} x (3)^{-1}}⁄{ 8 x 3^{3n} - 5 x (27)^{n}} We know 3^{3n} = (3^{3})^{n} = (27)^{n} Using this here, we get A = { 3 x (27)^{n} x (27) + 9 x (27)^{n} x (1⁄3}⁄{ 8 x (27)^{n} - 5 x (27)^{n}} We can see (27)^{n} being present in each term. Taking it common, we get A = [(27)^{n} { 3 x 27 + 9 x (1⁄3}]⁄[(27)^{n}{8 - 5 }] Cancelling (27)^{n} which is present in numerator and denominator, we get A = ( 81 + 3)⁄(3) = 84⁄3 = 28. Ans.

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Let A = (x^{1⁄3} - x^{-1⁄3})(x^{2⁄3} + 1 + x^{-2⁄3}) We know x^{2⁄3} = x^{2 x 1⁄3} = x^{(1⁄3) x 2} = {x^{(1⁄3)}}^{2} Similarly, x^{-2⁄3} = {x^{(-1⁄3)}}^{2} If we denote x^{1⁄3} by a, and x^{-1⁄3} by b, then ab = x^{1⁄3} x x^{-1⁄3} = 1 [Since p^{n} and p^{-n} are reciprocals to one another.] Thus A becomes (a - b)(a^{2} + ab + b^{2}). We know (a - b)(a^{2} + ab + b^{2}) = a^{3} - b^{3} ∴ A = a^{3} - b^{3} = (x^{1⁄3})^{3} - (x^{-1⁄3})^{3} = x^{1⁄3 x 3} - x^{-1⁄3 x 3} = x^{1} - x^{-1} = x - 1⁄x. Ans.

Solved Example 4 of Exponent Rules

Example 4 of Exponent Rules :

If abc = 1, prove that 1⁄(1 + a + b^{-1}) + 1⁄(1 + b + c^{-1}) + 1⁄(1 + c + a^{-1}) = 1.

Solution to Example 4 of Exponent Rules:

L.H.S. = 1⁄(1 + a + b^{-1}) + 1⁄(1 + b + c^{-1}) + 1⁄(1 + c + a^{-1}) = 1⁄(1 + a + 1⁄b) + 1⁄(1 + b + 1⁄c) + 1⁄(1 + c + 1⁄a) To prove, let us change the L.H.S. to two variables (from three), by using the relation abc = 1. Let us eliminate c. abc = 1 ⇒ c = 1⁄(ab); and c^{-1} = 1⁄c = ab; Substituting these in the L.H.S., we get L.H.S. = 1⁄(1 + a + 1⁄b) + 1⁄(1 + b + ab) + 1⁄(1 + 1⁄(ab) + 1⁄a) Multiplying the numerator and denominator of first term with b, of third term with ab, we get L.H.S. = (b x 1)⁄{b x (1 + a + 1⁄b)} + 1⁄(1 + b + ab) + (ab x 1)⁄{ab x (1 + 1⁄(ab) + 1⁄a)} = (b)⁄{b x 1 + b x a + b x (1⁄b)} + 1⁄(1 + b + ab) + (ab)⁄[ab x 1 + ab x {1⁄(ab)} + ab x (1⁄a)] = (b)⁄(b + ab + 1) + 1⁄(1 + b + ab) + (ab)⁄[ab + 1 + b ] = (b + 1 + ab)⁄(b + ab + 1) = 1 = R.H.S. (proved.)

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Simplify: {5 x (25)^{n+1} - 25 x (5)^{2n}}⁄{ 5 x 5^{(2n + 3)} - (25)^{n + 1}}

Solve: √(a⁄b) = (b⁄a)^{1 - 3x}

Solve: {(2⁄3)^{1⁄3}}^{(x - 1)} = 27⁄8.

For Answers, see at the bottom of the page.

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