There, we provided the explanation for Rational Exponents.

We applied the same 7 Laws and the 2 Rules in solving problems for Rational Exponents.

We provided a few solved examples and problems for practice with answers.

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

Studying the worked out problems will help remember and apply the 7 Laws and the 2 Rules for Rational Exponents.

Practice makes one perfect.

This is especially true for remembering Algebra Formulas (Math Formulas).

So, take the exercises seriously and practice solving the problems.

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

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If a^{x} = b^{y} = c^{z} and b⁄a = c⁄b, show that y⁄x = 2z⁄(x + z)

Solution to Example 1 of Fractional Exponents:

Let a^{x} = b^{y} = c^{z} = k ⇒ a = k^{1⁄x}; b = k^{1⁄y}; c = k^{1⁄z}; By data, b⁄a = c⁄b ⇒ (k^{1⁄y})⁄(k^{1⁄x}) = (k^{1⁄z})⁄(k^{1⁄y}) We know a^{m}⁄a^{n} = a^{m - n} Applying this here, we get (k^{1⁄y - 1⁄x}) = (k^{1⁄z - 1⁄y}) Since the bases are same, the exponents have to be equal. ∴ (1⁄y - 1⁄x) = (1⁄z - 1⁄y) ⇒ 1⁄y + 1⁄y = 1⁄z + 1⁄x ⇒ 2⁄y = (x + z)⁄xz Multiplying both sides with x⁄2, we get x⁄y = (x + z)⁄2z ⇒ y⁄x = 2z ⁄(x + z) (Proved.)

If y = 3^{1⁄3} + 1⁄(3^{1⁄3}), show that 3y^{3} - 9y = 10.

Solution to Example 2 of Fractional Exponents:

Let a = 3^{1⁄3}. Then a^{3} = 3........(i) and b = 1⁄(3^{1⁄3}). Then b^{3} = 1⁄3 .......(ii) Also, ab = 1.......(iii) and a + b = y........(iv). y = 3^{1⁄3} + 1⁄(3^{1⁄3}) = a + b y^{3} = (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b) Using (i), (ii), (iii) and (iv) here, we get y^{3} = 3 + 1⁄3 + 3(1)(y) Multiplying both sides with 3, we get 3y^{3} = 9 + 1 + 9y ⇒ 3y^{3} - 9y = 10. (Proved.)

Let a = x^{1⁄3}. Then x^{-1⁄3} = 1⁄a. The given equation becomes 2a + 2⁄a = 5. Multiplying both sides with a, we get 2a^{2} + 2 = 5a. ⇒ 2a^{2} - 5a + 2 = 0. This is a quadratic equation. We know

The solution of the quadratic equation ax^{2} + bx + c = 0 is x = {-b ± √( b^{2} - 4ac) }⁄2a

Applying this here, we get a = [-(-5) ± √{(-5)^{2} - 4(2)(2)}]⁄[2(2)] = [5 ± √{ 25 - 16 }]⁄[4] = [5 ± √{ 9 }]⁄[4] = [5 ± 3]⁄[4] = (5 + 3)⁄4 or (5 - 3)⁄4 = 8⁄4 or 2⁄4 = 2 or 1⁄2 . a = x^{1⁄3} ⇒ a^{3} = x; ∴ x = 2^{3} or (1⁄2)^{3} = 8 or 1⁄8. Ans.

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