FRACTIONAL EXPONENTS - SOLVED EXAMPLES AND EXERCISES ON RATIONAL EXPONENTS

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Please study  the Basics of Fractional Exponents,
if you have not already done so.

It is a prerequisite here.

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.

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Set of Solved Examples : Fractional Exponents

Solved Example 1 of Fractional Exponents

If ax = by = cz and ba = cb, show that yx = 2z⁄(x + z)

Solution to Example 1 of Fractional Exponents:

Let ax = by = cz = k
a = k1⁄x; b = k1⁄y; c = k1⁄z;
By data, ba = cb ⇒ (k1⁄y)⁄(k1⁄x) = (k1⁄z)⁄(k1⁄y)
We know aman = am - n
Applying this here, we get
(k1⁄y - 1⁄x) = (k1⁄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
xy = (x + z)⁄2zyx = 2z ⁄(x + z) (Proved.)





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Solved Example 2 of Fractional Exponents

If y = 31⁄3 + 1⁄(31⁄3), show that 3y3 - 9y = 10.

Solution to Example 2 of Fractional Exponents:

Let a = 31⁄3. Then a3 = 3........(i)
and b = 1⁄(31⁄3). Then b3 = 1⁄3 .......(ii)
Also, ab = 1.......(iii) and a + b = y........(iv).
y = 31⁄3 + 1⁄(31⁄3) = a + b
y3 = (a + b)3 = a3 + b3 + 3ab(a + b)
Using (i), (ii), (iii) and (iv) here, we get
y3 = 3 + 1⁄3 + 3(1)(y)
Multiplying both sides with 3, we get
3y3 = 9 + 1 + 9y ⇒ 3y3 - 9y = 10. (Proved.)

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Solved Example 3 of Fractional Exponents

Solve 2x1⁄3 + 2x-1⁄3 = 5.

Solution to Example 3 of Fractional Exponents:

Let a = x1⁄3. Then x-1⁄3 = 1⁄a.
The given equation becomes 2a + 2⁄a = 5.
Multiplying both sides with a, we get
2a2 + 2 = 5a. ⇒ 2a2 - 5a + 2 = 0.
This is a quadratic equation.
We know

The solution of the quadratic equation ax2 + bx + c = 0 is
x = {-b ± √( b2 - 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 = x1⁄3a3 = x;
x = 23 or (1⁄2)3 = 8 or 1⁄8. Ans.

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Exercise on Fractional Exponents

  1. Show that
    1⁄(1 + xa - b + xa - c) + 1⁄(1 + xb - c + xb - a) + 1⁄(1 + xc - a + xc - b) = 1.
  2. If ax = by = cz = dw and ab = cd,
    show that: 1⁄x + 1⁄y = 1⁄w + 1⁄z .
  3. If a1⁄3 + b1⁄3 + c1⁄3 = 0, show that: (a + b + c)3 = 27abc.

For Answers, see at the bottom of the page.

Answers to Exercise on Fractional Exponents

  1. (i) 1. (ii) 1.
  2. (i) 1⁄2. (ii) -8.

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