Chapter 09 — Powers — Part 2: Further Problems and Simplification of Indices

CHAPTER 09: POWERS

PART 2: FURTHER PROBLEMS AND SIMPLIFICATION OF INDICES

In Part 1, we studied zero, negative and fractional indices. We also learnt how to convert between surds and fractional index notation.

In this lesson, we develop those ideas further. We shall combine several laws of indices, simplify complicated numerical and algebraic expressions, work with fractions and brackets, and solve equations involving powers.

The main objective is to develop accuracy and confidence when dealing with expressions containing different types of indices.

1. Revising the Laws of Indices

The laws of indices apply to expressions involving the same base. They remain valid for fractional and negative indices whenever the expressions are defined.

Multiplication

a^m × aⁿ = a^m+n

Division

a^m ÷ aⁿ = a^m−n

Power of a power

(a^m)ⁿ = a^mn

Power of a product

(ab)ⁿ = aⁿbⁿ

Power of a quotient

(a/b)ⁿ = aⁿ/bⁿ

Remember these three operations:

• Multiplying like bases: add the indices.
• Dividing like bases: subtract the indices.
• Raising a power to another power: multiply the indices.

2. Simplifying Expressions by Combining Several Laws

Many examination questions require more than one law of indices. The best approach is to simplify the numerical coefficients separately from the variables, then combine the indices.

Example 1

Simplify:

x³ × x⁻² × x⁴

Solution

All the terms have the same base, x. Add their indices.

= x³⁻²⁺⁴

= x⁵

Example 2

Simplify:

a^7/3 × a^−2/3 ÷ a^1/3

Solution

Add the indices for multiplication and subtract the index for division.

= a^{7/3 − 2/3 − 1/3}

= a^4/3

= a^4/3

Example 3

Simplify:

(x²)³ × x⁻⁴

Solution

First, multiply the indices inside the power.

= x⁶ × x⁻⁴

Then add the indices.

= x⁶⁻⁴

= x²

3. Simplifying Algebraic Expressions with Coefficients

When an expression contains numbers and variables, simplify the numerical coefficients and the powers of each variable separately.

Example 4

Simplify:

3x²y × 4x³y⁻²

Solution

Multiply the numerical coefficients:

3 × 4 = 12

Combine the powers of x:

x² × x³ = x⁵

Combine the powers of y:

y × y⁻² = y⁻¹

Therefore:

= 12x⁵y⁻¹

Expressing the answer using positive indices:

= 12x⁵/y

Example 5

Simplify:

12a⁵b⁻² ÷ 3a⁻¹b

Solution

Divide the coefficients:

12 ÷ 3 = 4

Subtract the indices of a:

a⁵⁻⁽⁻¹⁾ = a⁶

Subtract the indices of b:

b⁻²⁻¹ = b⁻³

Hence:

= 4a⁶b⁻³

= 4a⁶/b³

4. Powers of Brackets

When an expression inside brackets is raised to a power, the outside index applies to every factor inside the brackets.

(ab)ⁿ = aⁿbⁿ

Example 6

Simplify:

(2x³y²)²

Solution

Apply the outside index to every factor.

= 2² × (x³)² × (y²)²

= 4x⁶y⁴

= 4x⁶y⁴

Example 7

Simplify:

(8x⁶)^2/3

Solution

Apply the fractional index to both factors.

= 8^2/3 × x^6×2/3

= (∛8)² × x⁴

= 4x⁴

= 4x⁴

Example 8

Simplify:

(27a⁶b³)^−1/3

Solution

Apply the negative fractional index to each factor.

= 27^−1/3 × a⁻² × b⁻¹

= 1/(3a²b)

= 1/(3a²b)

5. Simplifying Fractions Involving Indices

When a fraction contains powers of the same variable, divide the coefficients and subtract the indices.

Example 9

Simplify:

(15x⁷y⁴)/(5x²y)

Solution

= (15/5)x⁷⁻²y⁴⁻¹

= 3x⁵y³

= 3x⁵y³

Example 10

Simplify and express using positive indices:

(6a⁻²b³)/(3a²b⁻¹)

Solution

= 2a⁻²⁻²b³⁻⁽⁻¹⁾

= 2a⁻⁴b⁴

= 2b⁴/a⁴

6. Fractions with a Sum or Difference in the Numerator

A fraction can sometimes be separated into simpler fractions when its numerator contains a sum or difference and its denominator is a single term.

For example:

(x² + 3x)/x = x²/x + 3x/x

= x + 3

Example 11

Simplify:

(12ab⁻² − 16a)/(8ab)

Solution

Separate the numerator terms:

= 12ab⁻²/(8ab) − 16a/(8ab)

Simplify each fraction:

= (3/2)b⁻³ − 2b⁻¹

Express using positive indices:

= 3/(2b³) − 2/b

Important warning:

You may separate the terms in the numerator when the denominator is a single term. However:

1/(x + y) ≠ 1/x + 1/y

A sum in the denominator cannot be separated in this way.

7. Expressing Answers Using Positive Indices

In many examination questions, you may be asked to express an answer using positive indices only.

Remember:

a⁻ⁿ = 1/aⁿ

Example 12

Simplify and express using positive indices:

x⁻²y³ ÷ x⁻⁵y⁻¹

Solution

= x^{−2−(−5)}y³⁻⁽⁻¹⁾

= x³y⁴

Answer: x³y⁴

Example 13

Simplify:

(a⁻²b³)⁻²

Solution

Multiply each index by −2.

= a⁴b⁻⁶

Write using positive indices:

= a⁴/b⁶

8. Solving Equations Involving Indices

When both sides of an equation can be expressed using the same positive base, their indices can be equated.

Example 14

Solve:

2^x+1 = 16

Solution

Express 16 as a power of 2.

2^x+1 = 2⁴

Equate the indices:

x + 1 = 4

x = 4 − 1

x = 3

Example 15

Solve:

9^x = 27

Solution

Write both sides using base 3.

(3²)^x = 3³

3^2x = 3³

Equate the indices:

2x = 3

x = 3/2

Example 16

Solve:

125^2x−1 = 5^x+6

Solution

Express 125 as a power of 5.

(5³)^2x−1 = 5^x+6

5^6x−3 = 5^x+6

Equate the indices:

6x − 3 = x + 6

5x = 9

x = 9/5

9. Finding an Unknown Index

Sometimes a question gives an expression involving an unknown index and asks you to find its value.

Example 17

Find k if:

(2³)^k = 2¹²

Solution

2^3k = 2¹²

3k = 12

k = 4

Example 18

Find p if:

(x^p × x^1/2)/x⁻³ = x⁶

Assume x is positive and non-zero, and the identity is required to hold for all such x.

Solution

Combine the indices on the left.

x^{p+1/2−(−3)} = x⁶

x^p+1/2+3 = x⁶

Equate the indices:

p + 1/2 + 3 = 6

p + 7/2 = 6

p = 6 − 7/2

p = 5/2

10. Common Examination Mistakes

1. Adding indices when dividing. Division requires subtraction.
2. Adding instead of multiplying in a power of a power. For example, (x³)⁴ = x¹², not x⁷.
3. Forgetting the numerical coefficient. In (3x²)³, the 3 must also be cubed.
4. Leaving negative indices when positive indices are requested. Convert them into reciprocals.
5. Equating indices with different bases. First express both sides using the same base.
6. Separating a denominator containing a sum. A fraction such as 1/(x + y) cannot be split into 1/x + 1/y.
7. Ignoring restrictions. Variables appearing in denominators cannot be zero.

11. Exercises

Attempt the following questions independently. Show your working before opening each solution.

Question 1

Simplify:

(a) x⁴ × x⁻³ × x⁵

(b) a⁷/³ × a⁻²/³

(c) y⁵ ÷ y⁻²

Show worked solutions

(a) x⁴⁻³⁺⁵ = x⁶

(b) a^7/3−2/3 = a^5/3

(c) y⁵⁻⁽⁻²⁾ = y⁷

Question 2

Simplify:

(a) 3x²y × 5x³y⁻²

(b) 12a⁵b⁻² ÷ 4a²b

(c) 8m⁻³n² × 2m⁵n⁻⁴

Show worked solutions

(a) 15x⁵y⁻¹ = 15x⁵/y

(b) 3a³b⁻³ = 3a³/b³

(c) 16m²n⁻² = 16m²/n²

Question 3

Simplify:

(a) (x³)⁴

(b) (2a²b³)²

(c) (27x⁶)¹/³

Show worked solutions

(a) x^3×4 = x¹²

(b) 2²a⁴b⁶ = 4a⁴b⁶

(c) 27¹/³ × x² = 3x²

Question 4

Simplify and express using positive indices:

(a) x⁻³y²

(b) a⁻²b⁴ ÷ a³b⁻¹

(c) (m⁻²n³)⁻²

Show worked solutions

(a) y²/x³

(b) a⁻²⁻³b⁴⁻⁽⁻¹⁾ = a⁻⁵b⁵ = b⁵/a⁵

(c) m⁴n⁻⁶ = m⁴/n⁶

Question 5

Simplify:

(16x⁸)^3/4

Show worked solution

= 16^3/4x^8×3/4

= 8x⁶

Answer: 8x⁶

Question 6

Simplify:

(81a⁴b⁸)^−1/4

Show worked solution

= 81^−1/4a⁻¹b⁻²

= 1/(3ab²)

Answer: 1/(3ab²)

Question 7

Simplify:

(18x⁵y²)/(6x⁻¹y⁴)

Show worked solution

= 3x⁵⁻⁽⁻¹⁾y²⁻⁴

= 3x⁶y⁻²

= 3x⁶/y²

Question 8

Simplify:

(12ab⁻² − 8a)/(4ab)

Show worked solution

= 12ab⁻²/(4ab) − 8a/(4ab)

= 3b⁻³ − 2b⁻¹

= 3/b³ − 2/b

Question 9

Solve:

4^x+1 = 64

Show worked solution

4^x+1 = 4³

x + 1 = 3

x = 2

Question 10

Solve:

25^2x−1 = 5^x+4

Show worked solution

(5²)^2x−1 = 5^x+4

5^4x−2 = 5^x+4

4x − 2 = x + 4

3x = 6

x = 2

Question 11

Find k if:

(3²)^k × 3⁻² = 3⁶

Show worked solution

3^2k × 3⁻² = 3⁶

3^2k−2 = 3⁶

2k − 2 = 6

2k = 8

k = 4

Question 12

Given that y = 3x⁻², express the following in the form kxⁿ, where k and n are constants:

y²/(6x⁻¹)

Show worked solution

Substitute y = 3x⁻².

= (3x⁻²)²/(6x⁻¹)

= 9x⁻⁴/(6x⁻¹)

= (3/2)x^{−4−(−1)}

= (3/2)x⁻³

Therefore k = 3/2 and n = −3.

Question 13

Simplify and express the answer as a single power of 5:

(5^1/4)² × (5^2/3)^−3/4 ÷ (5⁻¹)⁻²

Show worked solution

= 5^1/2 × 5^−1/2 ÷ 5²

= 5^{1/2−1/2−2}

= 5⁻²

Question 14

A student simplifies (x³)⁴ as x⁷. Explain the error and give the correct answer.

Show worked solution

The student has added the indices instead of multiplying them.

For a power raised to another power, multiply the indices:

(x³)⁴ = x^3×4

= x¹²

12. Lesson Summary

Multiplication: add indices.

a^m × aⁿ = a^m+n

Division: subtract indices.

a^m ÷ aⁿ = a^m−n

Power of a power: multiply indices.

(a^m)ⁿ = a^mn

Negative index: take the reciprocal.

a⁻ⁿ = 1/aⁿ

Fractional index: use the corresponding root.

a^m/n = (ⁿ√a)^m

After completing this lesson, you should be able to:

• Combine several laws of indices in one expression.
• Simplify numerical and algebraic expressions involving powers.
• Apply fractional and negative indices to brackets.
• Simplify algebraic fractions involving indices.
• Express answers using positive indices.
• Solve equations by expressing both sides using the same base.
• Find unknown indices in algebraic identities.

END OF PART 2

You have completed further problems and simplification of indices.

Next: Part 3 — Surds and Their Simplification.

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