Chapter 10 — Part 5: Revision, Mixed Examination Questions and Consolidation of Logarithms

CHAPTER 10 — PART 5

Revision, Mixed Examination Questions and Consolidation of Logarithms

Introduction

In the previous parts of this chapter, we studied logarithms from their definition through to their applications. We have seen how logarithms simplify expressions, help us solve equations involving unknown powers, and transform certain non-linear relationships into straight-line forms.

This final part brings these ideas together. It is designed to help you select an appropriate method when faced with an unfamiliar examination question.

The aim is not simply to memorise formulae. You should understand why a method works, show each important step, and check whether your final answers satisfy the original conditions.

1. Essential Revision of Logarithms

1.1 Meaning of a logarithm

A logarithm answers the question: to what power must a given base be raised to obtain a particular positive number?

If a^x = N, then log_aN = x.

For a real logarithm, the conditions are:

• The base must be positive: a > 0.
• The base must not equal one: a ≠ 1.
• The argument must be positive: N > 0.

Example 1

Evaluate log₂32.

We ask: 2 raised to which power gives 32?

Since 2⁵ = 32,

Therefore, log₂32 = 5.

1.2 The fundamental laws

For positive numbers M and N, and a valid base a:

Product law

log_aM + log_aN = log_a(MN)

Quotient law

log_aM − log_aN = log_a(M/N)

Power law

k log_aM = log_a(M^k)

Important: A logarithm does not distribute over addition or subtraction. In general, log_a(M + N) is not equal to log_aM + log_aN.

1.3 Useful special results

log_a1 = 0

log_aa = 1

log_a(a^x) = x

a^log_ax = x, for x > 0

1.4 Change of base

A logarithm can be evaluated using another convenient base. This is particularly useful when a calculator provides only common logarithms or natural logarithms.

log_aN = log_bN / log_ba

In particular:

log_aN = ln N / ln a

Example 2

Evaluate log₃20 correct to three decimal places.

Using change of base:

log₃20 = ln 20 / ln 3

= 2.995732... / 1.098612...

Therefore, log₃20 ≈ 2.727.

2. Combining and Simplifying Logarithms

Before solving a logarithmic equation, it is often useful to combine separate logarithmic terms into one logarithm. The laws of logarithms allow us to do this systematically.

Example 3

Express the following as a single logarithm:

2 log₃x + log₃5 − log₃y

First apply the power law:
2 log₃x = log₃x².

Apply the product law:
log₃x² + log₃5 = log₃(5x²).

Apply the quotient law:

log₃(5x²/y)

Answer: log₃(5x²/y).

The original logarithms require x > 0 and y > 0.

Example 4

Expand fully:

log₅(25x³/√y)

Separate the numerator and denominator:
log₅25 + log₅x³ − log₅√y.

Apply the power law:
2 + 3 log₅x − ½ log₅y.

Answer: 2 + 3 log₅x − ½ log₅y.

Conditions: x > 0 and y > 0.

3. Solving Logarithmic Equations

A logarithmic equation contains the unknown in one or more logarithmic expressions. The usual strategy is to combine logarithms, convert to an algebraic equation, and then check the original domain restrictions.

Remember: Every logarithmic argument must be positive. A value obtained by algebra is not automatically a valid solution.

Example 5

Solve:

log₂(x + 1) = 3

Convert to exponential form:
x + 1 = 2³.

x + 1 = 8.

x = 7.

Answer: x = 7.

Check: x + 1 = 8, which is positive. The solution is valid.

Example 6

Solve:

log₃x + log₃(x − 2) = 1

Step 1: State the restrictions.

x > 0 and x − 2 > 0.
Therefore, x > 2.

Step 2: Combine the logarithms.

log₃[x(x − 2)] = 1

Step 3: Convert to exponential form.

x(x − 2) = 3¹ = 3

x² − 2x − 3 = 0.

(x − 3)(x + 1) = 0.

x = 3 or x = −1.

Step 4: Check the restrictions.

The restriction is x > 2. Therefore, x = −1 is not acceptable. Only x = 3 is valid.

Answer: x = 3.

Example 7

Solve:

log₁₀(x + 4) − log₁₀(x − 2) = 1

Step 1: Restrictions.

x + 4 > 0 and x − 2 > 0.
Hence x > 2.

Step 2: Apply the quotient law.

log₁₀((x + 4)/(x − 2)) = 1

Step 3: Convert to exponential form.

(x + 4)/(x − 2) = 10

x + 4 = 10x − 20.

24 = 9x.

x = 8/3.

Since 8/3 > 2, the solution satisfies the restriction.

Answer: x = 8/3.

4. Equations Involving Quadratic Expressions in Logarithms

Some equations become quadratic after a substitution. If the same logarithmic expression appears more than once, it may be helpful to represent it by a single variable.

Example 8

Solve:

(log₂x)² − 5log₂x + 6 = 0

Step 1: Make a substitution.

Let u = log₂x.

Then u² − 5u + 6 = 0.

Step 2: Factorise.

(u − 2)(u − 3) = 0

Therefore, u = 2 or u = 3.

Step 3: Return to x.

log₂x = 2 gives x = 2² = 4.

log₂x = 3 gives x = 2³ = 8.

Answers: x = 4 or x = 8.

Example 9

Solve:

log₃(x)² − 4log₃x + 3 = 0

Here, the notation means (log₃x)².

Let u = log₃x.

u² − 4u + 3 = 0.

(u − 1)(u − 3) = 0.

u = 1 or u = 3.

log₃x = 1 gives x = 3.

log₃x = 3 gives x = 27.

Answers: x = 3 or x = 27.

5. Logarithmic Inequalities

When solving inequalities involving logarithms, we must pay attention to the base. The logarithmic function is increasing when its base is greater than one and decreasing when its base lies between zero and one.

Case 1: Base greater than one

If a > 1, the logarithmic function is increasing. The inequality direction is preserved when comparing logarithms with the same base.

log_aM > log_aN ⇔ M > N

Case 2: Base between zero and one

If 0 < a < 1, the logarithmic function is decreasing. The inequality direction is reversed.

log_aM > log_aN ⇔ M < N

Example 10

Solve:

log₂(x + 1) > 2

Domain: x + 1 > 0, so x > −1.

Since 2 > 1, the logarithmic function is increasing.

x + 1 > 2².

x + 1 > 4.

x > 3.

Answer: x > 3.

Example 11

Solve:

log_1/2(x − 1) > 2

Domain: x − 1 > 0, so x > 1.

Since 0 < 1/2 < 1, the logarithmic function is decreasing. Reverse the inequality when converting:

x − 1 < (1/2)²

x − 1 < 1/4.

x < 5/4.

Combining this with x > 1 gives:

Answer: 1 < x < 5/4.

Examination reminder: Always combine the inequality obtained from the logarithm with the original domain restriction. Do not forget the domain.

6. Solving Exponential Equations Using Logarithms

An exponential equation has the unknown appearing in the exponent. If the two sides cannot conveniently be expressed using the same base, logarithms can be used to bring the exponent down.

If a^x = b, then x = log_ab

Example 12

Solve:

5^x = 17

Take natural logarithms on both sides:

ln(5^x) = ln 17

x ln 5 = ln 17.

x = ln 17 / ln 5.

Answer: x ≈ 1.760.

Example 13

Solve:

3^{2x + 1} = 7^{x − 2}

Take natural logarithms on both sides:

(2x + 1)ln 3 = (x − 2)ln 7

2x ln 3 + ln 3 = x ln 7 − 2ln 7.

2x ln 3 − x ln 7 = −2ln 7 − ln 3.

x(2ln 3 − ln 7) = −2ln 7 − ln 3.

x = (−2ln 7 − ln 3)/(2ln 3 − ln 7)

Answer: x ≈ 6.42.

Example 14

Solve:

2e^3x = 11

Divide both sides by 2:
e^3x = 11/2.

Take natural logarithms:
3x = ln(11/2).

x = ln(11/2)/3.

Answer: x ≈ 0.568.

7. Linearisation of Relationships

Linearisation is the process of transforming a relationship that is not linear into a straight-line form. This is useful when estimating unknown constants from experimental or observed data.

7.1 Power relationships

Consider the relationship:

y = axⁿ

Taking natural logarithms:

ln y = ln a + n ln x

Compare this with the straight-line equation y = mx + c:

• Vertical variable: ln y.
• Horizontal variable: ln x.
• Gradient: n.
• Vertical intercept: ln a.
• Therefore, a = e^intercept.

Example 15

A relationship is given by y = axⁿ. The following observations are made:

x	y
1	3
2	12
3	27

Determine a and n.

When x = 1, y = 3:
3 = a(1)ⁿ, so a = 3.

Use x = 2 and y = 12:
12 = 3(2)ⁿ.

4 = 2ⁿ, so n = 2.

Therefore, a = 3, n = 2 and y = 3x².

The third observation confirms the relationship: 3(3)² = 27.

7.2 Exponential relationships

Consider:

y = kb^x

Taking natural logarithms gives:

ln y = ln k + x ln b

Comparing with a straight line:

• Vertical variable: ln y.
• Horizontal variable: x.
• Gradient: ln b.
• Vertical intercept: ln k.
• Therefore, b = e^gradient.
• And k = e^intercept.

Example 16

A relationship is modelled by y = kb^x. It is known that y = 6 when x = 0 and y = 24 when x = 2. Find k and b.

When x = 0:
6 = kb⁰ = k.
Therefore, k = 6.

When x = 2:
24 = 6b².

b² = 4.

Since the exponential base is positive, b = 2.

Therefore, k = 6, b = 2 and y = 6(2)^x.

7.3 Recognising the correct linear form

Original relationship	Linear form	Gradient
y = axn	ln y = ln a + n ln x	n
y = kbx	ln y = ln k + x ln b	ln b
y = Aekx	ln y = ln A + kx	k

For logarithmic linearisation, the quantities whose logarithms are taken must be positive. Check that the model and the supplied data allow this.

8. Exponential Growth and Decay

Exponential models are used when the rate of change of a quantity is proportional to the quantity itself. They are useful in modelling population growth, continuous compound growth, and certain decay processes.

8.1 Exponential growth

A = A₀e^kt,   k > 0

Where:

• A is the quantity after time t.
• A₀ is the initial quantity.
• k is the positive growth constant.
• t is the elapsed time.

8.2 Exponential decay

A = A₀e^−kt,   k > 0

The negative exponent represents a decrease in the quantity over time.

Example 17 — Growth

A population initially contains 500 organisms. It grows according to the model:

P = 500e^0.04t

Find the population after 10 time units.

Substitute t = 10:

P = 500e^0.04(10)

P = 500e^0.4.

P ≈ 500 × 1.491825.

P ≈ 746 organisms.

Example 18 — Decay

A quantity has an initial value of 800 units and follows the model:

A = 800e^−0.15t

Find the amount remaining after 4 time units.

Substitute t = 4:

A = 800e^−0.15(4)

A = 800e^−0.6.

A ≈ 800 × 0.548812.

A ≈ 439.05 units.

8.3 Finding the time taken

Logarithms are particularly useful when the unknown is the time in an exponential model.

Example 19

A quantity decreases according to:

A = 600e^−0.08t

How long does it take to decrease to 300 units?

Substitute A = 300:

300 = 600e^−0.08t

Divide by 600:
1/2 = e^−0.08t.

Take natural logarithms:
ln(1/2) = −0.08t.

t = ln(1/2)/(−0.08).

Answer: t ≈ 8.664 time units.

8.4 Half-life

The half-life is the time required for a decaying quantity to reduce to half of its initial value.

For the model A = A₀e^−kt, at half-life T:

1/2 = e^−kT

Taking natural logarithms:

T = ln 2 / k

Example 20

A substance follows the decay model A = A₀e^−0.12t. Find its half-life.

Here k = 0.12.

T = ln 2 / 0.12.

T ≈ 5.776 time units.

9. Mixed Examination-Style Worked Questions

The questions in this section combine different ideas. In an examination, the first task is often to recognise which mathematical method is required.

Example 21

Solve the equation:

log₂(x + 3) + log₂(x − 1) = 3

Step 1: Establish the domain.

x + 3 > 0 and x − 1 > 0.
Therefore, x > 1.

Step 2: Combine the logarithms.

log₂[(x + 3)(x − 1)] = 3

Step 3: Convert to exponential form.

(x + 3)(x − 1) = 8

x² + 2x − 3 = 8.

x² + 2x − 11 = 0.

Step 4: Solve the quadratic.

x = [−2 ± √(4 + 44)]/2

x = [−2 ± √48]/2.

x = −1 ± 2√3.

Step 5: Apply the domain restriction.

Since x > 1, the negative root is rejected.

Answer: x = −1 + 2√3.

Example 22

Solve:

4^x − 5(2^x) + 4 = 0

Step 1: Express the terms using one base.

Since 4^x = (2^x)², we have:

(2^x)² − 5(2^x) + 4 = 0

Step 2: Substitute.

Let u = 2^x.

u² − 5u + 4 = 0

(u − 1)(u − 4) = 0.

u = 1 or u = 4.

Step 3: Return to x.

2^x = 1 gives x = 0.

2^x = 4 gives x = 2.

Answers: x = 0 or x = 2.

Example 23

A quantity is modelled by:

N = 120e^0.06t

Find the time taken for the quantity to reach 200.

Put N = 200:

200 = 120e^0.06t

Divide by 120:
5/3 = e^0.06t.

Take natural logarithms:
ln(5/3) = 0.06t.

t = ln(5/3)/0.06.

Answer: t ≈ 8.514 time units.

10. Common Errors to Avoid

1. Adding inside a logarithm incorrectly.
   Do not write log(M + N) = log M + log N.
2. Ignoring the domain.
   Every logarithmic argument must be positive.
3. Forgetting to check roots.
   An algebraic root may not satisfy the original logarithmic equation.
4. Reversing an inequality incorrectly.
   A logarithm with base between zero and one is decreasing.
5. Confusing the two linearisation methods.
   For y = axⁿ, plot ln y against ln x. For y = kb^x, plot ln y against x.
6. Using the wrong sign in a decay model.
   For decay, write A = A₀e^−kt with k positive.
7. Rounding too early.
   Keep sufficient calculator accuracy until the final answer.

11. Student Practice Exercises

Attempt the following questions independently before opening the answer sections. Show all your working, especially where domain restrictions or logarithm laws are involved.

Exercise A: Laws and Evaluation

1. Evaluate log₂64.
2. Evaluate log₅(1/25).
3. Express 3log₄x − log₄y as a single logarithm.
4. Expand log₃(9x²/√y) fully.
5. Evaluate log₇20 correct to three decimal places.

Show answers to Exercise A

1. 6.
2. −2.
3. log₄(x³/y).
4. 2 + 2log₃x − ½log₃y.
5. Approximately 1.540.

Exercise B: Logarithmic Equations

1. Solve log₂(x − 3) = 4.
2. Solve log₅x + log₅(x − 4) = 1.
3. Solve log₃(x + 2) − log₃x = 1.
4. Solve (log₂x)² − 3log₂x + 2 = 0.
5. Solve log₁₀(x + 1) + log₁₀(x − 1) = 1.

Show answers to Exercise B

1. x = 19.
2. x = 5.
3. x = 1.
4. x = 2 or x = 4.
5. x = √11, taking the positive root required by the domain.

Exercise C: Inequalities

1. Solve log₃(x − 2) > 2.
2. Solve log_1/3(x + 1) > 1.
3. Solve log₂(2x − 1) ≤ 3.

Show answers to Exercise C

1. x > 11.
2. −1 < x < −2/3.
3. 1/2 < x ≤ 9/2.

Exercise D: Exponential Equations

1. Solve 3^x = 20, correct to three decimal places.
2. Solve 2e^4x = 15.
3. Solve 5^{x + 1} = 2^2x.
4. Solve 9^x − 10(3^x) + 9 = 0.

Show answers to Exercise D

1. x = ln 20 / ln 3 ≈ 2.727.
2. x = ln(15/2)/4 ≈ 0.504.
3. x = ln 5/(2ln 2 − ln 5) ≈ −1.756.
4. x = 0 or x = 2.

Exercise E: Linearisation and Modelling

1. A relationship is y = axⁿ. Given that y = 8 when x = 2 and y = 27 when x = 3, determine a and n.
2. A relationship is y = kb^x. Given y = 5 when x = 0 and y = 45 when x = 2, determine k and b.
3. A population follows P = 400e^0.03t. Find P after 20 time units.
4. A quantity follows A = 900e^−0.1t. Find the time taken for it to decrease to 300.
5. A decay model is A = A₀e^−0.2t. Find the half-life.

Show answers to Exercise E

1. a = 1 and n = 3.
2. k = 5 and b = 3.
3. P = 400e^0.6 ≈ 729 units.
4. t = ln 3/0.1 ≈ 10.986 time units.
5. T = ln 2/0.2 ≈ 3.466 time units.

12. Final Mixed Examination Questions

These questions are longer and require you to decide which methods to combine. They are intended to test understanding rather than memorisation.

Question 1

(a) State the conditions required for log_ax to be defined.

(b) Solve: log₂(x + 5) + log₂(x − 1) = 3.

(c) Explain why any root that does not satisfy the original logarithmic restrictions must be rejected.

Show solution to Question 1

(a) a > 0, a ≠ 1 and x > 0.

(b) First, x > 1.

log₂[(x + 5)(x − 1)] = 3

(x + 5)(x − 1) = 8

x² + 4x − 5 = 8.

x² + 4x − 13 = 0.

x = −2 ± √17.

Since x > 1, only x = −2 + √17 is valid.

Answer: x = −2 + √17.

(c) A logarithm of zero or a negative number is not defined in the real number system. Such a root cannot satisfy the original equation.

Question 2

The relationship between two positive variables x and y is y = axⁿ.

(a) Show that the relationship can be written in a straight-line form using natural logarithms.

(b) State the gradient and vertical intercept of the resulting line when ln y is plotted against ln x.

(c) If the gradient is 2 and the vertical intercept is ln 4, determine a and n.

Show solution to Question 2

(a)

y = axⁿ

ln y = ln a + n ln x

(b)

Gradient = n; vertical intercept = ln a.

(c)

Since the gradient is 2, n = 2. Since the intercept is ln 4, ln a = ln 4, so a = 4.

Therefore, y = 4x².

Question 3

A quantity decreases according to the model A = 1000e^−0.05t.

(a) Find the amount remaining after 12 time units.

(b) Find the time taken for the quantity to fall to 250.

(c) Find the half-life of the quantity.

Show solution to Question 3

(a)

A = 1000e^−0.05(12)

A ≈ 548.812 units.

(b)

250 = 1000e^−0.05t

1/4 = e^−0.05t.

ln(1/4) = −0.05t.

t = ln(1/4)/(−0.05).

t ≈ 27.726 time units.

(c)

T = ln 2/0.05

T ≈ 13.863 time units.

13. Chapter 10 Revision Checklist

Before moving to the next chapter, make sure you can confidently do the following:

• Convert between exponential and logarithmic forms.
• State the restrictions on logarithmic bases and arguments.
• Apply the product, quotient and power laws correctly.
• Expand and combine logarithmic expressions.
• Use the change-of-base formula.
• Solve logarithmic equations and check all solutions.
• Solve equations involving quadratic expressions in logarithms.
• Solve logarithmic inequalities with the correct direction.
• Solve exponential equations using logarithms.
• Transform power and exponential relationships into linear form.
• Interpret gradients and intercepts in logarithmic graphs.
• Use exponential models for growth and decay.
• Calculate the time taken for a quantity to reach a given value.
• Calculate half-life from an exponential decay model.

14. Chapter Summary

Logarithms are the inverse operations of exponentiation. Their laws allow products, quotients and powers to be handled more conveniently. They are also essential tools for solving equations in which the unknown appears in an exponent.

In linearisation, logarithms transform certain power and exponential relationships into straight-line equations. This makes it possible to estimate unknown constants from data.

Exponential models describe many processes in which the rate of change is proportional to the amount present. Logarithms allow us to calculate the time required to reach a specified amount.

The most important habits in this chapter are to choose the correct logarithm law, respect domain restrictions, show clear working, and check the final answer in the original equation.

End of Chapter 10 — Logarithms

You have now completed the five parts of this chapter, from the foundations of logarithms to equations, inequalities, linearisation, exponential models and mixed examination practice.

Continue revising the worked examples and attempt the exercises again without looking at the solutions. Confidence comes from understanding each step and being able to reproduce it independently.

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