Chapter 10: Logarithms — Part 2: Change of Base and Further Applications

CHAPTER 10: LOGARITHMS

PART 2: CHANGE OF BASE, EVALUATING LOGARITHMS AND FURTHER APPLICATIONS

Mathematics HipHop — A-Level Pure Mathematics

In Part 1, we introduced logarithms, converted between logarithmic and exponential forms, and established the fundamental laws of logarithms.

In this part, we extend those ideas. We investigate how to evaluate logarithms with unfamiliar bases, how to change the base of a logarithm, and how logarithms can be used to solve equations involving unknown powers.

Careful algebraic working is essential. A calculator may help with numerical evaluation, but it does not replace mathematical reasoning.

1. Learning objectives

By the end of this lesson, you should be able to:

• Explain and derive the change-of-base formula.
• Evaluate logarithms using common logarithms or natural logarithms.
• Use a calculator to obtain logarithmic values to a specified degree of accuracy.
• Apply reciprocal and chain relationships between logarithms.
• Solve equations involving logarithms.
• Solve exponential equations by taking logarithms of both sides.
• Check solutions against the restrictions of the original equations.

2. Common logarithms and natural logarithms

Although logarithms can have many different bases, two bases are especially common in mathematics.

(a) Common logarithms

A common logarithm is a logarithm to base 10. The base is usually omitted in its notation.

log x = log₁₀x

For example:

log 1000 = 3

because 10³ = 1000.

(b) Natural logarithms

A natural logarithm is a logarithm to base e, where e is an irrational mathematical constant approximately equal to 2.71828.

ln x = log_ex

For example:

ln e = 1

because e¹ = e.

Both common and natural logarithms follow the same laws as logarithms to any other valid base.

Worked Example 1

Evaluate the following:

1. log 10000
2. log 0.01
3. ln e⁴
4. ln 1

Reveal the complete solution

1. Since 10⁴ = 10000:
   log 10000 = 4
2. Since 10⁻² = 0.01:
   log 0.01 = −2
3. Using ln(eˣ) = x:
   ln e⁴ = 4
4. Since e⁰ = 1:
   ln 1 = 0

3. The change-of-base formula

In Part 1, we evaluated logarithms whose bases could be handled directly using known powers. However, what happens when we need to evaluate log₇20?

Most scientific calculators provide keys for common logarithms and natural logarithms. The change-of-base formula allows us to express a logarithm in terms of another suitable base.

The change-of-base formula

log_aN = log_bN / log_ba

Here, a and b must be positive and different from 1, and N must be positive.

In particular:

log_aN = log N / log a

log_aN = ln N / ln a

Derivation of the change-of-base formula

Let:

x = log_aN

Writing this in exponential form gives:

aˣ = N

Take logarithms to base b on both sides:

log_b(aˣ) = log_bN

Apply the power law:

x log_ba = log_bN

Divide both sides by log_ba:

x = log_bN / log_ba

But x = log_aN. Therefore:

log_aN = log_bN / log_ba

Hence, the change-of-base formula is established.

Worked Example 2

Express log₅17 in terms of common logarithms.

Reveal the complete solution

Using the change-of-base formula:

log_aN = log N / log a

Therefore:

log₅17 = log 17 / log 5

Answer:

log₅17 = log 17 / log 5

4. Evaluating logarithms using a calculator

The change-of-base formula makes it possible to evaluate logarithms with unfamiliar bases using a scientific calculator.

Worked Example 3

Evaluate log₇20 correct to four significant figures.

Reveal the complete solution

Apply the change-of-base formula:

log₇20 = log 20 / log 7

Using a scientific calculator:

log 20 ≈ 1.301030

log 7 ≈ 0.845098

Therefore:

log₇20 ≈ 1.539

Answer: 1.539 correct to four significant figures.

Check the reasonableness of the answer: 7¹ = 7 and 7² = 49. Since 20 lies between 7 and 49, its logarithm to base 7 must lie between 1 and 2.

Worked Example 4

Evaluate log₃10 correct to three decimal places.

Reveal the complete solution

Using natural logarithms:

log₃10 = ln 10 / ln 3

Substitute calculator values:

= 2.302585... / 1.098612...

≈ 2.096

Answer: 2.096 correct to three decimal places.

Calculator reminder:

When entering the change-of-base formula, use brackets around the numerator and denominator:

(log 20) ÷ (log 7)

This helps prevent errors in calculator entry.

5. Reciprocal relationships between logarithms

The change-of-base formula leads to a useful relationship between logarithms whose bases and arguments are interchanged.

log_ab = 1 / log_ba

This is called the reciprocal relationship.

Explanation

Using the change-of-base formula:

log_ab = log b / log a

Similarly:

log_ba = log a / log b

Taking the reciprocal of the second expression gives:

1 / log_ba = log b / log a

Therefore:

log_ab = 1 / log_ba

Worked Example 5

Given that log₂5 = p, express log₅2 in terms of p.

Reveal the complete solution

Using the reciprocal relationship:

log₅2 = 1 / log₂5

But log₂5 = p.

log₅2 = 1/p

Answer: 1/p

6. The chain rule for logarithms

Another useful relationship is obtained when the argument of one logarithm is the base of another.

log_ab × log_bc = log_ac

Provided all bases and arguments satisfy the conditions for real logarithms.

Derivation

Using the change-of-base formula:

log_ab = log b / log a

log_bc = log c / log b

Multiplying:

(log b / log a)(log c / log b)

Cancel the common factor log b:

= log c / log a

Therefore:

log_ab × log_bc = log_ac

Worked Example 6

Evaluate:

log₂3 × log₃8

Reveal the complete solution

Apply the chain rule:

log₂3 × log₃8 = log₂8

Since 2³ = 8:

= 3

Answer: 3

Worked Example 7

Simplify:

log₂5 × log₅7 × log₇16

Reveal the complete solution

Apply the chain rule successively:

= log₂7 × log₇16

= log₂16

Since 2⁴ = 16:

= 4

Answer: 4

7. Expressing logarithms in terms of given quantities

In examination questions, you may be given the values of certain logarithms and asked to express another logarithm in terms of those values.

Worked Example 8

Given that log_a2 = m and log_a3 = n, express log_a72 in terms of m and n.

Reveal the complete solution

First express 72 as a product of powers of 2 and 3:

72 = 2³ × 3²

Therefore:

log_a72 = log_a(2³ × 3²)

Apply the product law:

= log_a2³ + log_a3²

Apply the power law:

= 3log_a2 + 2log_a3

Substitute the given values:

= 3m + 2n

Answer: 3m + 2n

Worked Example 9

Given that log_a2 = p and log_a5 = q, express log_a20 in terms of p and q.

Reveal the complete solution

Factorise 20:

20 = 2² × 5

Take logarithms:

log_a20 = log_a(2² × 5)

Apply the product and power laws:

= 2log_a2 + log_a5

Substitute:

= 2p + q

Answer: 2p + q

8. Solving equations involving logarithms

Logarithmic equations can often be solved by combining logarithmic laws and then converting the resulting equation into exponential form.

Important procedure:

1. State the restrictions that make the logarithms defined.
2. Use the logarithmic laws to combine or simplify terms.
3. Convert the logarithmic equation into exponential form where appropriate.
4. Solve the resulting algebraic equation.
5. Check that every solution satisfies the original restrictions.

Worked Example 10

Solve:

log₂x = 5

Reveal the complete solution

Convert into exponential form:

x = 2⁵

x = 32

The value is positive, so it satisfies the logarithm's domain restriction.

Answer: x = 32

Worked Example 11

Solve:

log₃(x + 1) = 2

Reveal the complete solution

First, the argument must be positive:

x + 1 > 0

x > −1

Convert to exponential form:

x + 1 = 3²

x + 1 = 9

x = 8

Since 8 is greater than −1, it is valid.

Answer: x = 8

Worked Example 12

Solve:

log₂x + log₂(x − 2) = 3

Reveal the complete solution

Step 1: State the restrictions.

Both arguments must be positive:

x > 0 and x − 2 > 0

Therefore x > 2

Step 2: Apply the product law.

log₂[x(x − 2)] = 3

Step 3: Convert to exponential form.

x(x − 2) = 2³

x² − 2x = 8

Step 4: Rearrange into a quadratic equation.

x² − 2x − 8 = 0

(x − 4)(x + 2) = 0

x = 4 or x = −2

Step 5: Apply the restriction.

Since x must be greater than 2, x = −2 is rejected.

Answer: x = 4

9. Solving exponential equations using logarithms

Sometimes an unknown appears as an index and the equation cannot easily be expressed using the same base. Logarithms provide a method for solving such equations.

General method

Suppose:

aˣ = N

Taking logarithms to a valid base b:

log_b(aˣ) = log_bN

Apply the power law:

x log_ba = log_bN

Therefore:

x = log_bN / log_ba

Worked Example 13

Solve 3ˣ = 20, giving your answer correct to three significant figures.

Reveal the complete solution

Take logarithms to base 10:

log(3ˣ) = log 20

Apply the power law:

x log 3 = log 20

Divide by log 3:

x = log 20 / log 3

x ≈ 2.727...

Answer: x = 2.73 to three significant figures.

Worked Example 14

Solve:

5^2x−1 = 12

Give your answer correct to three significant figures.

Reveal the complete solution

Take logarithms of both sides:

log(5^2x−1) = log 12

Apply the power law:

(2x − 1)log 5 = log 12

Divide by log 5:

2x − 1 = log 12 / log 5

Rearrange:

2x = 1 + (log 12 / log 5)

x = [1 + (log 12 / log 5)]/2

x ≈ 1.272...

Answer: x = 1.27 to three significant figures.

10. Applications of logarithms to linear relationships

Logarithms can transform certain non-linear relationships into straight-line equations. This is useful when studying mathematical models and determining unknown constants.

Case A: A power relationship

Suppose two variables are related by:

y = kxⁿ

Take logarithms of both sides:

log y = log(kxⁿ)

Apply the product and power laws:

log y = log k + n log x

Linear form:

Y = C + nX

where Y = log y, X = log x and C = log k.

A graph of log y against log x is a straight line with:

• Gradient = n
• Vertical intercept = log k

Case B: An exponential relationship

Suppose:

y = kaˣ

Taking logarithms:

log y = log(kaˣ)

log y = log k + x log a

Linear form:

Y = C + mx

where Y = log y, C = log k and m = log a.

A graph of log y against x is a straight line with:

• Gradient = log a
• Vertical intercept = log k

Worked Example 15

A relationship between two positive variables is given by:

y = kxⁿ

It is known that log y = 0.8 when log x = 1, and log y = 2.0 when log x = 4. Find n and k.

Reveal the complete solution

From the logarithmic form:

log y = log k + n log x

The two points are (1, 0.8) and (4, 2.0).

Step 1: Find the gradient.

n = (2.0 − 0.8)/(4 − 1)

n = 1.2/3 = 0.4

Step 2: Find log k.

Substitute log x = 1 and log y = 0.8:

0.8 = log k + 0.4(1)

log k = 0.4

Step 3: Find k.

k = 10^0.4

k ≈ 2.512

Answers:

n = 0.4,   k ≈ 2.512

11. Common examination mistakes

1. Using the change-of-base formula incorrectly: The denominator must contain the logarithm of the original base.
2. Forgetting brackets: Enter the entire numerator and denominator correctly when using a calculator.
3. Confusing common and natural logarithms: log usually means base 10, while ln means base e.
4. Reversing the reciprocal relationship: log_ab = 1/log_ba.
5. Combining different bases directly: The product and quotient laws require the same base.
6. Ignoring domain restrictions: Every logarithmic argument must be positive.
7. Keeping invalid roots: Check all solutions in the original logarithmic equation.
8. Rounding too early: Keep sufficient calculator accuracy until the final answer.
9. Misidentifying the gradient: In log y = log k + n log x, the gradient is n when plotting log y against log x.

12. Student exercises

Attempt the questions independently before revealing the answers. Give numerical answers to the stated accuracy.

Exercise A: Common and natural logarithms

1. Evaluate log 100000. [1 mark]
2. Evaluate log 0.0001. [2 marks]
3. Evaluate ln(e⁶). [1 mark]
4. Evaluate ln(1/e³). [2 marks]
5. Write log_e15 using the standard logarithmic notation. [1 mark]

Exercise B: Change of base

1. Express log₄13 in terms of common logarithms. [2 marks]
2. Express log₇25 in terms of natural logarithms. [2 marks]
3. Evaluate log₅18 correct to three significant figures. [3 marks]
4. Evaluate log₆50 correct to three decimal places. [3 marks]

Exercise C: Relationships between logarithms

1. Given that log₂7 = p, express log₇2 in terms of p. [2 marks]
2. Evaluate log₂5 × log₅32. [2 marks]
3. Simplify log₃2 × log₂5 × log₅27. [3 marks]
4. Given that log_a2 = m and log_a7 = n, express log_a28 in terms of m and n. [3 marks]

Exercise D: Logarithmic equations

1. Solve log₂x = 6. [2 marks]
2. Solve log₅(x − 1) = 2. [2 marks]
3. Solve log₃x + log₃(x − 2) = 1. [4 marks]
4. Solve log₂(x + 1) = 3. [2 marks]

Exercise E: Exponential equations and applications

1. Solve 2ˣ = 15, correct to three significant figures. [3 marks]
2. Solve 7^2x = 20, correct to three decimal places. [3 marks]
3. Solve 4^x+1 = 30, correct to three significant figures. [3 marks]
4. Given y = kxⁿ, explain what the gradient represents when log y is plotted against log x. [2 marks]

13. Answers to exercises

Exercise A — Answers

1. 5
2. −4
3. 6
4. −3
5. ln 15

Exercise B — Answers

1. log 13 / log 4
2. ln 25 / ln 7
3. log₅18 ≈ 1.80
4. log₆50 ≈ 2.183

Exercise C — Answers

1. 1/p
2. 5
3. 3
4. 2m + n

Exercise D — Answers and working

1. x = 64
2. x − 1 = 25, so x = 26.

3. First, x > 2. Apply the product law:

   log₃[x(x − 2)] = 1
   x(x − 2) = 3
   x² − 2x − 3 = 0
   (x − 3)(x + 1) = 0

   x = 3 or x = −1. Since x > 2, the valid answer is x = 3.

4. x + 1 = 8, so x = 7.

Exercise E — Answers and working

1. x = log 15 / log 2

   x ≈ 3.91 to three significant figures.

2. 2x log 7 = log 20
   x = log 20/(2 log 7)

   x ≈ 0.770 to three decimal places.

3. (x + 1)log 4 = log 30
   x = log 30/log 4 − 1

   x ≈ 1.45 to three significant figures.

4. The gradient is n, the index of x in y = kxⁿ.

14. Part 2 revision checklist

Before moving to Part 3, make sure you can confidently:

• Distinguish between common and natural logarithms.
• Derive and apply the change-of-base formula.
• Evaluate logarithms using a scientific calculator.
• Use reciprocal and chain relationships.
• Express logarithms in terms of given quantities.
• Solve logarithmic equations while checking restrictions.
• Solve exponential equations using logarithms.
• Transform power and exponential relationships into linear forms.
• Interpret gradients and intercepts in logarithmic graphs.

Final message to the learner

Logarithms provide a powerful way of working with unknown powers. The change-of-base formula extends the range of logarithms we can evaluate, while the reciprocal and chain relationships make complicated expressions easier to handle.

When solving equations, always remember that a mathematically obtained answer must also satisfy the conditions of the original logarithmic expression.

Practise consistently and show every important step in your working.

MATHEMATICS HIPHOP — MAKING MATHEMATICS EASIER, ONE CONCEPT AT A TIME.