DP Maths AA SL - Number and Algebra - Exponential and Logarithmic Functions

DP Maths AA SL - Number and Algebra - Exponential and Logarithmic Functions - Full notes

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DP Maths AA SL | Exponential and Logarithmic Functions

Exponential and Logarithmic Functions

This topic contains 1 note:

1.5 Exponents and Logarithms

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.5

Exponents and Logarithms

Full notes for SL · Syllabus SL 1.5, 1.7

Contents

1	Laws of exponents
2	Rational exponents
3	Surds and rationalizing denominators
4	Introducing logarithms
5	Laws of logarithms
6	Change of base
7	Solving exponential equations
8	Solving logarithmic equations
9	Applications: exponential models and logarithmic scales
10	Equations with the same base
11	Equations that become quadratics
12	Logarithmic equations with several terms
13	Proving the laws of logarithms
14	Exponential models and logarithms
15	Graphs of exponential and logarithmic functions
16	Mixed review: worked problems
17	Additional exam-style practice
18	Exam-style questions with markscheme
19	Common misconceptions and links
20	Check your understanding
21	Summary and topic map
22	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
apply the laws of exponents, including zero, negative and rational exponents
simplify surds and rationalize denominators
understand a logarithm as an exponent and convert between exponential and logarithmic form
prove and use the laws of logarithms
use the change of base formula
solve exponential equations by matching bases, taking logarithms, or substitution
solve logarithmic equations and reject invalid solutions
apply exponents and logarithms to models and logarithmic scales

This booklet covers SL 1.5, 1.7, which is common to SL and HL.

Guiding questions

How are exponents and logarithms connected? How can logarithms solve equations where the unknown is in the exponent? Why do scientists measure some quantities on logarithmic scales?

How this topic is examined

Exponents and logarithms underpin much of the course: sequences (finding n), financial mathematics, exponential functions, calculus and statistics. Paper 1 tests exact manipulation (simplify, write in the form a + logb c, solve by matching bases or a substitution); Paper 2 tests modelling problems solved with technology.

Show the law you are using at each step, write exact answers in the form requested, and check that solutions of logarithmic equations make every argument positive.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Laws of exponents

Rules for multiplying, dividing and raising powers.

Think about it

210 = 1024, which is about a thousand, and 220 is about a million. Without a calculator, roughly what is 230? Why?

In the expression an, a is the base and n is the exponent (index or power). For a positive integer n, an means n factors of a multiplied together. The laws of exponents follow directly from counting factors.

a3 × a2 = (a × a × a)(a × a) = a5: multiplying powers of the same base adds exponents. a5 ÷ a2 = a3: dividing subtracts exponents. (a3)2 = a3 × a3 = a6: a power of a power multiplies exponents.

The definitions of zero and negative exponents are chosen so that these laws keep working. Since a3 ÷ a3 = 1 but the law gives a0, we define a0 = 1. Since a0 ÷ an = 1/an but the law gives a−n, we define a−n = 1/an. A negative exponent means a reciprocal, not a negative number.

WORKED EXAMPLE 1 Simplifying (basic)

Simplify (a) x7 × x−3; (b) (3y2)3; (c) 12p5 ÷ 4p−2.

Solution

(a) x7 + (−3) = x4A1

(b) 33y6 = 27y6A1

WORKED EXAMPLE 2 Several laws together (standard)

Simplify (2x3y−1)3 ÷ (4x5y2), giving your answer with positive exponents.

Solution

(2x3y−1)3 = 8x9y−3M1

8x9y−3 ÷ 4x5y2 = 2x4y−5M1

= 2x4/y5A1

WORKED EXAMPLE 3 Writing in a single base (exam level)

Write 4x × 8x−1 ÷ 16 as a single power of 2.

Solution

4x = 22x, 8x−1 = 23x−3, 16 = 24M1

22x + 3x − 3 − 4M1

= 25x−7A1

Common mistake

Writing am × an = amn, or multiplying powers with different bases by adding exponents: 23 × 32 ≠ 65. The laws only apply to the same base (or the same exponent, for (ab)n).

Practice exercise 1

1. Simplify x9 ÷ x−2.

2. Evaluate 5−2 and (2/3)−3.

3. Simplify (5a2b)2 × 2a−3b4.

4. Simplify (x−2y3)−2.

5. Write 9x+1 × 272x as a power of 3.

6. Simplify (6x4y−2)/(9x−1y3).

7. Explain, using the laws of exponents, why 70 = 1.

2 Rational exponents

Fractional powers are roots.

What should a1/2 mean? The power law requires (a1/2)2 = a1 = a, so a1/2 must be the number whose square is a: a1/2 = √a. In the same way, a1/n is the nth root of a, because (a1/n)n = a.

For a general rational exponent m/n, am/n = (a1/n)m: take the nth root, then raise to the power m. It is usually easier to take the root first, because the numbers stay small: 272/3 = (∛27)2 = 32 = 9.

WORKED EXAMPLE 4 Evaluating rational powers (basic)

Without a calculator, evaluate (a) 323/5; (b) 125−2/3.

Solution

(a) (fifth root of 32)3 = 23 = 8A1

(b) 1/(cube root of 125)2 = 1/52 = 1/25A1

WORKED EXAMPLE 5 A fraction to a negative rational power (standard)

Evaluate (16/81)−3/4.

Solution

(16/81)−3/4 = (81/16)3/4M1

= (3/2)3M1

= 27/8A1

WORKED EXAMPLE 6 Simplifying with rational exponents (exam level)

Write (√x × x2) ÷ ∛x in the form xk.

Solution

x1/2 × x2 ÷ x1/3M1

k = 1/2 + 2 − 1/3 = 13/6M1

x13/6A1

Practice exercise 2

1. Evaluate 813/4.

2. Evaluate 8−4/3.

3. Write 1/√x3 as a power of x.

4. Simplify (x1/2 + x−1/2)2.

5. Solve x3/2 = 64.

6. Write ∛(x2) × √x as a single power.

3 Surds and rationalizing denominators

Exact values with roots.

A surd is an irrational root such as √2, √5 or ∛3. Surds are kept exact on Paper 1 instead of being rounded. Two rules let us simplify them: √(ab) = √a × √b and √(a/b) = √a/√b. To simplify √72, find the largest square factor: √72 = √(36 × 2) = 6√2.

Like surds can be added: 3√2 + 5√2 = 8√2, but √2 + √3 cannot be simplified. Note that √(a + b) ≠ √a + √b.

To rationalize a denominator means to rewrite a fraction so the denominator has no surd. For a single surd, multiply top and bottom by that surd. For a denominator of the form a + √b, multiply by its conjugate a − √b, because (a + √b)(a − √b) = a2 − b is rational (difference of two squares).

WORKED EXAMPLE 7 Simplifying surds (basic)

Simplify √50 + √18 − √8.

Solution

5√2 + 3√2 − 2√2M1

= 6√2A1

WORKED EXAMPLE 8 Expanding brackets (standard)

Expand and simplify (3 + 2√5)(1 − √5).

Solution

3 − 3√5 + 2√5 − 2 × 5M1

= −7 − √5A1

WORKED EXAMPLE 9 Rationalizing with a conjugate (exam level)

Write (5 + √3)/(2 − √3) in the form a + b√3.

Solution

Multiply by (2 + √3)/(2 + √3)M1

Numerator: 10 + 5√3 + 2√3 + 3 = 13 + 7√3; denominator: 4 − 3 = 1A1

= 13 + 7√3A1

Common mistake

Writing √(9 + 16) = 3 + 4. In fact √25 = 5. The root of a sum is not the sum of the roots.

Practice exercise 3

1. Simplify √75 − √12.

2. Rationalize 6/√2.

3. Expand (√7 + √3)(√7 − √3).

4. Write 4/(3 + √5) in the form a + b√5.

5. Simplify (1 + √2)2.

6. Show that 1/(√2 + 1) + 1/(√2 − 1) = 2√2.

4 Introducing logarithms

A logarithm is an exponent.

Think about it

You know that 23 = 8. How would you answer the question 'what power of 2 gives 8?' without guessing, if the number were 10 instead of 8?

The logarithm of b to base a is the exponent to which a must be raised to give b. In other words, logarithms answer the question 'what power?':

So log2 8 = 3 because 23 = 8; log10 0.01 = −2 because 10−2 = 0.01; log9 3 = 1/2 because 91/2 = 3.

Two bases are used most often. Base 10, the common logarithm, is written log x (or lg x). Base e, where e ≈ 2.718 28, is the natural logarithm, written ln x. Your calculator has keys for both.

Because ax is always positive, logarithms are only defined for positive arguments: loga 0 and loga(−5) do not exist. The base must also be positive and not equal to 1.

Logarithms and exponentials are inverse functions: each undoes the other. So loga ax = x, and a raised to the power loga x equals x. In particular loga 1 = 0 (because a0 = 1) and loga a = 1.

Exponential form	Logarithmic form
25 = 32	log2 32 = 5
103 = 1000	log 1000 = 3
e0 = 1	ln 1 = 0
4−1/2 = 1/2	log4(1/2) = −1/2

WORKED EXAMPLE 10 Converting forms (basic)

(a) Write 53 = 125 in logarithmic form. (b) Find x if log3 x = −2. (c) Write ln y = 4 in exponential form.

Solution

(a) log5 125 = 3A1

(b) x = 3−2 = 1/9A1

WORKED EXAMPLE 11 Evaluating without a calculator (standard)

Evaluate (a) log4 32; (b) log1/3 27; (c) ln(1/√e).

Solution

(a) 4x = 32: 22x = 25, x = 5/2A1

(b) (1/3)x = 27: 3−x = 33, x = −3A1

Practice exercise 4

1. Evaluate log2(1/16).

2. Evaluate log8 4.

3. Write ln x = 3 in exponential form.

4. Find x if logx 125 = 3.

5. Evaluate 10log 7 and e2 ln 3.

6. Explain why log2(−4) is undefined.

5 Laws of logarithms

Logs turn products into sums.

Because logarithms are exponents, the laws of exponents give corresponding laws of logarithms (for x, y > 0):

Why it works

Let loga x = p and loga y = q, so x = ap and y = aq. Then xy = apaq = ap+q, so loga(xy) = p + q = loga x + loga y. Similarly x/y = ap−q gives the quotient law, and xm = (ap)m = amp gives loga xm = mp = m loga x.

These laws let us combine several logarithms into one, or expand one logarithm into several. Historically, they made multiplication of large numbers possible by adding logarithms from printed tables.

WORKED EXAMPLE 12 Combining into a single logarithm (basic)

Write 2 log 5 + log 12 − log 3 as a single logarithm and evaluate.

Solution

= log 25 + log 12 − log 3M1

= log(25 × 12 ÷ 3) = log 100M1

= 2A1

WORKED EXAMPLE 13 Expanding (standard)

Expand ln(x3√y/z2).

Solution

ln x3 + ln y1/2 − ln z2M1

= 3 ln x + ½ ln y − 2 ln zA1

WORKED EXAMPLE 14 In terms of given logs (exam level)

Given loga 2 = p and loga 3 = q, write (a) loga 18 and (b) loga(8a/9) in terms of p and q.

Solution

(a) loga(2 × 32) = p + 2qA1

(b) loga 8 + loga a − loga 9M1

= 3p + 1 − 2qA1

Common mistake

Writing log(x + y) = log x + log y, or (log x)/(log y) = log(x/y). Neither is a law of logarithms.

Practice exercise 5

1. Simplify log3 54 − log3 2.

2. Write 3 log 2 + log 5 − log 4 as a single logarithm.

3. Expand log(100x2/√y).

4. Given ln 2 = a and ln 5 = b, write ln 0.4 in terms of a and b.

5. Show that loga 1/x = −loga x.

6. Simplify (log 27)/(log 9).

6 Change of base

Convert any logarithm into one your calculator can evaluate.

Why it works

Let y = loga x, so ay = x. Take logs to base b of both sides: logb ay = logb x, so y logb a = logb x. Dividing gives y = logb x / logb a.

Usually b = 10 or b = e. For example, log5 20 = ln 20 ÷ ln 5 = 1.86. A useful special case is loga b = 1 ÷ logb a.

WORKED EXAMPLE 15 Change of base (standard)

Show that log9 x = ½ log3 x, and hence solve log3 x + log9 x = 6.

Solution

log9 x = log3 x ÷ log3 9 = ½ log3 xM1 AG

log3 x + ½ log3 x = 6, so log3 x = 4M1

x = 81A1

Practice exercise 6

1. Evaluate log7 50 to 3 s.f.

2. Show that log4 x = (log2 x)/2.

3. Solve log2 x = log4(x + 12).

4. Simplify log2 3 × log3 8.

7 Solving exponential equations

Match bases, take logs, or substitute.

An exponential equation has the unknown in an exponent. There are three main methods.

1. Matching bases (exact, Paper 1): write both sides as powers of the same base, then equate exponents. 4x+1 = 8x: 22x+2 = 23x, so 2x + 2 = 3x and x = 2.

2. Taking logarithms: when bases cannot be matched, take logs of both sides and use the power law to bring the unknown down. 3x = 20 gives x ln 3 = ln 20, so x = ln 20 / ln 3 = 2.73.

3. Substitution (quadratic in disguise): equations such as 4x − 5 × 2x + 4 = 0 become quadratics when we let y = 2x (since 4x = (2x)2 = y2): y2 − 5y + 4 = 0, so y = 1 or y = 4, giving x = 0 or x = 2. Reject any value of y that is negative or zero, because 2x > 0.

WORKED EXAMPLE 16 Matching bases (basic)

Solve 9x = 27x−1.

Solution

32x = 33x−3M1

2x = 3x − 3, x = 3A1

WORKED EXAMPLE 17 Exact answer with logs (standard)

Solve 52x−1 = 3x, giving x in the form ln a / ln b.

Solution

(2x − 1) ln 5 = x ln 3M1

x(2 ln 5 − ln 3) = ln 5M1

x = ln 5 / ln(25/3)A1

WORKED EXAMPLE 18 A quadratic in disguise (exam level)

Solve 22x+1 − 9 × 2x + 4 = 0.

Solution

Let y = 2x: 22x+1 = 2y2, so 2y2 − 9y + 4 = 0M1

(2y − 1)(y − 4) = 0: y = ½ or y = 4A1

x = −1 or x = 2A1 A1

WORKED EXAMPLE 19 An equation with e (exam level)

Solve ex − 6e−x = 1.

Solution

Multiply by ex: e2x − ex − 6 = 0M1

(ex − 3)(ex + 2) = 0A1

ex = −2 has no solution, so x = ln 3R1 A1

Figure 1. y = 2x and y = log2 x are reflections of each other in y = x.

Practice exercise 7

1. Solve 8x = 32.

2. Solve 2x = 7 to 3 s.f.

3. Solve 3x+2 = 5x exactly.

4. Solve 9x − 4 × 3x + 3 = 0.

5. Solve e2x = 5ex − 6.

6. Solve 2x × 32x = 100 to 3 s.f.

8 Solving logarithmic equations

Combine, convert, check.

To solve a logarithmic equation: use the laws to combine the logarithms into a single logarithm on each side; convert to exponential form (or equate arguments if both sides are single logs to the same base); solve; and check every solution in the original equation, rejecting any that make an argument zero or negative.

WORKED EXAMPLE 20 Combining logs (standard)

Solve log2 x + log2(x − 2) = 3.

Solution

log2 x(x − 2) = 3, so x(x − 2) = 8M1

x2 − 2x − 8 = 0, (x − 4)(x + 2) = 0A1

x = 4; x = −2 is rejected because log2(−2) is undefinedR1 A1

WORKED EXAMPLE 21 Logs on both sides (exam level)

Solve 2 ln x − ln(x + 2) = ln 3.

Solution

ln(x2/(x + 2)) = ln 3, so x2 = 3x + 6M1

x2 − 3x − 6 = 0, x = (3 ± √33)/2A1

x = (3 + √33)/2; the negative root is rejectedR1

Key point

Always check solutions of logarithmic equations: rejecting invalid roots is usually worth a reasoning mark.

Practice exercise 8

1. Solve ln x = 2 ln 3.

2. Solve log(x + 1) − log x = 1.

3. Solve log3(2x + 1) = 2.

4. Solve log6 x + log6(x − 5) = 2.

5. Solve (ln x)2 = 3 ln x.

9 Applications: exponential models and logarithmic scales

Why logs appear in science.

Logarithms solve for time in exponential models: if a population grows as P = 500 × 1.08t, the time to reach 2000 is t = ln 4 / ln 1.08 = 18.0 years.

Logarithmic scales compress huge ranges of values. The pH of a solution is −log[H+]; each pH unit is a tenfold change. Sound intensity in decibels is 10 log(I/I0); an increase of 10 dB means ten times the intensity. The Richter scale for earthquakes is logarithmic: magnitude 7 is ten times the amplitude of magnitude 6.

WORKED EXAMPLE 22 Comparing on a log scale (exam level)

Earthquake magnitude M = log(A/A0). One earthquake has magnitude 6.2 and another 7.5. How many times larger is the amplitude of the second?

Solution

A = A0 × 10MM1

Ratio = 107.5 − 6.2 = 101.3M1

= 20.0 timesA1

Practice exercise 9

1. How long does it take $1000 to double at 5% compound interest per year?

2. A sound increases from 60 dB to 85 dB. By what factor does the intensity increase?

3. The pH of a solution is 3.4. Find [H+].

10 Equations with the same base

Equate the exponents.

If both sides can be written as powers of the same base, the exponents must be equal: ap = aq implies p = q (a > 0, a ≠ 1). Rewrite numbers such as 8, 16, 1/4 and √2 as powers of 2, or 9, 27 and 1/3 as powers of 3.

WORKED EXAMPLE 23 Same base (standard)

Solve 8x+1 = 42x.

Solution

23x+3 = 24xM1

3x + 3 = 4x: x = 3A1

WORKED EXAMPLE 24 Fractional powers (exam level)

Solve 9x = 27√3.

Solution

32x = 33 × 31/2 = 37/2M1

x = 7/4A1

Practice exercise 10

1. Solve 25x = 125.

2. Solve 2x × 4x+1 = 32.

3. Solve (1/3)x = 81.

11 Equations that become quadratics

Substitute for the exponential.

Equations such as 4x − 3 × 2x − 4 = 0 or e2x − 5ex + 6 = 0 are quadratics in 2x or ex. Substitute y = 2x (so 4x = y2), solve, and reject any negative value of y, since exponentials are always positive.

WORKED EXAMPLE 25 A quadratic in a power of 2 (exam level)

Solve 4x − 3(2x) − 4 = 0.

Solution

y2 − 3y − 4 = 0 with y = 2xM1

(y − 4)(y + 1) = 0; y = −1 rejectedA1 R1

2x = 4: x = 2A1

WORKED EXAMPLE 26 A quadratic in an exponential (exam level)

Solve e2x − 5ex + 6 = 0.

Solution

(ex − 2)(ex − 3) = 0M1

x = ln 2 or ln 3A1 A1

Practice exercise 11

1. Solve 9x − 10(3x) + 9 = 0.

2. Solve ex − 6e−x = 1.

3. Solve 22x+1 − 9(2x) + 4 = 0.

12 Logarithmic equations with several terms

Combine first, check the domain.

Use the laws of logarithms to combine terms into a single log, convert to exponential form, solve, and reject any solution that makes an argument of a log non-positive. Logs with different bases must first be converted to a common base.

WORKED EXAMPLE 27 Two log terms (exam level)

Solve log2(x + 2) + log2(x − 4) = 4.

Solution

(x + 2)(x − 4) = 16M1

x2 − 2x − 24 = 0: x = 6 or −4A1

x = 6 (x = −4 rejected)R1 A1

WORKED EXAMPLE 28 Different bases (exam level)

Solve log3 x + log9 x = 6.

Solution

log9 x = ½ log3 xM1

1.5 log3 x = 6: log3 x = 4A1

x = 81A1

Practice exercise 12

1. Solve ln x + ln(x + 3) = ln 10.

2. Solve 2 log x − log(x − 2) = log 8.

3. Solve log4(x + 6) = 1 + log4 x.

13 Proving the laws of logarithms

From the laws of exponents.

Why it works

Let loga x = p and loga y = q, so x = ap and y = aq. Product: xy = ap+q, so loga(xy) = p + q. Quotient: x/y = ap−q. Power: xm = amp, so loga xm = mp = m loga x. Change of base: if y = loga x then ay = x; taking logb of both sides gives y logb a = logb x.

WORKED EXAMPLE 29 Using the laws (standard)

Given loga 2 = p and loga 3 = q, express loga 72 and loga(4/27) in terms of p and q.

Solution

72 = 23 × 32: 3p + 2qA1

4/27: 2p − 3qA1

Practice exercise 13

1. Given log 5 = k, express log 2 in terms of k.

2. Show that loga b × logb a = 1.

14 Exponential models and logarithms

Solving for time.

In models N = N0ekt or N = N0at, logarithms find the time at which a level is reached, the growth constant from two data points, and doubling or half-life times.

WORKED EXAMPLE 30 Finding k from data (standard)

N = N0ekt. N(0) = 300 and N(5) = 750. Find k and N(12).

Solution

750 = 300e5k: k = ln(2.5)/5 = 0.183M1 A1

N(12) = 300e12k = 2701A1

WORKED EXAMPLE 31 Comparing two models (exam level)

Town A: 20 000(1.03)t; town B: 30 000(1.01)t. When will A overtake B?

Solution

(1.03/1.01)t = 1.5M1

t = ln 1.5/ln(1.03/1.01) = 20.7 yearsA1

Practice exercise 14

1. A substance decays with half-life 12 years. Find k in N = N0e−kt.

2. An investment grows from $5000 to $8000 in 6 years with continuous growth. Find the rate.

15 Graphs of exponential and logarithmic functions

Inverse pairs.

y = ax and y = loga x are inverse functions: their graphs are reflections in y = x. Transformations move the asymptotes: y = 2x − 3 has asymptote y = −3; y = ln(x + 2) has asymptote x = −2.

Figure 2. Exponential and logarithmic functions are inverses.

WORKED EXAMPLE 32 Intercepts and asymptotes (standard)

For y = 3 − 2e−x, find the intercepts and the asymptote.

Solution

y-intercept (0, 1)A1

x-intercept: e−x = 1.5, x = −ln 1.5 = −0.405A1

Asymptote y = 3A1

Practice exercise 15

1. Find the domain, asymptote and x-intercept of y = log3(x − 1).

2. Find the inverse of f(x) = e2x + 1.

16 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 33 Mixed problem 1 (exam level)

Solve 32x+1 = 5x, giving the answer exactly.

Solution

(2x + 1) ln 3 = x ln 5M1

x(2 ln 3 − ln 5) = −ln 3A1

x = −ln 3/ln(9/5)A1

WORKED EXAMPLE 34 Mixed problem 2 (exam level)

Simplify (log2 27)(log3 16).

Solution

(3 log2 3)(4 log3 2)M1

= 12 log2 3 log3 2 = 12A1

Practice exercise 16

1. Solve ex + 2e−x = 3.

2. Solve 2x = 3x−1.

3. Simplify log4 8 + log9 27.

17 Additional exam-style practice

Questions in the style of Papers 1 and 2.

WORKED EXAMPLE 35 Paper 1 question (exam level)

(a) Write 2 ln 3 + ln 4 − ln 6 as a single logarithm. (b) Hence solve ey = 2 ln 3 + ln 4 − ln 6 exactly.

Solution

(a) ln(9 × 4/6) = ln 6M1 A1

(b) ey = ln 6: y = ln(ln 6)A1

WORKED EXAMPLE 36 Paper 2 question (exam level)

The pH of a solution is −log[H+]. Lemon juice has pH 2.3 and milk pH 6.6. How many times greater is the hydrogen ion concentration of lemon juice?

Solution

106.6 − 2.3M1

= 104.3 = 19 950A1

Practice exercise 17

1. Solve log(x + 1) + log(x − 1) = 1.

2. Find x if 52x = 4 × 5x + 5.

18 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style, no calculator) (about 15 minutes)

(a) Write log3 45 − 2 log3 5 + log3(5/3) in the form log3 k. [2]

(b) Solve 22x+1 − 9(2x) + 4 = 0. [4]

Markscheme

Part	Answer	Marks
(a)	log3(45 × (5/3) ÷ 25)	M1
	log3 3 (k = 3)	A1
(b)	2y2 − 9y + 4 = 0 with y = 2x	M1
	(2y − 1)(y − 4) = 0	A1
	y = ½ or 4	A1
	x = −1 or x = 2	A1
(c)	(x + 3)(x − 3) = 16	M1
	x2 = 25	A1
	x = ±5	A1
	x = 5 only (x = −5 gives negative arguments)	R1

Examiner's comments. In (b), candidates must recognize 22x+1 = 2(2x)2. In (c), the reasoning mark is lost if x = −5 is not explicitly rejected.

Question 2 (Paper 1 style, no calculator) (about 12 minutes)

Let a = log2 3.

(a) Write log2 12 in terms of a. [2]

(b) Write log4 27 in terms of a. [3]

Markscheme

Part	Answer	Marks
(a)	log2 4 + log2 3	M1
	2 + a	A1
(b)	log2 27 / log2 4	M1
	3a/2	A1 A1
(c)	x = log4 27	M1
	x = 3a/2	A1

Examiner's comments. Part (c) uses 'hence': linking back to (b) earns full marks quickly.

19 Common misconceptions and links

Students often think…	Actually…
log(x + y) = log x + log y	only log(xy) = log x + log y
a−n is negative	a−n = 1/an
all solutions of log equations are valid	check arguments are positive
√(a + b) = √a + √b	only √(ab) = √a√b

How 1.5 connects to the rest of the course

Topic	Connection
1.1 Sequences	solving for n
1.4 Financial applications	time to reach a target
2.8 Exponential and log functions	graphs
5.6 Differentiation rules	derivatives of ex and ln x

20 Check your understanding

Section A: multiple choice

1. 272/3 =

A. 9 B. 18 C. 3 D. 6

2. log2 32 =

A. 16 B. 4 C. 5 D. 6

3. ln e4 =

A. e4 B. 4 C. 1 D. 0

4. log a + log b =

A. log(a + b) B. log a × log b C. log(a/b) D. log ab

5. log5 20 ≈

A. 4 B. 1.86 C. 0.54 D. 2.30

6. If 3x = 81, x =

A. 4 B. 27 C. 3 D. 9

Section B: structured

7. Simplify (√50 − √8)/√2. [2]

8. Solve e2x − 3ex + 2 = 0. [3]

9. Prove loga b × logb a = 1. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. (5√2 − 2√2)/√2 = 3.

8. ex = 1 or 2: x = 0 or ln 2.

9. Change of base: (ln b/ln a)(ln a/ln b) = 1.

21 Summary and topic map

Idea	Remember
Exponent laws	add, subtract, multiply exponents; a0 = 1; a−n = 1/an
Rational exponents	am/n = (nth root of a)m
Surds	simplify square factors; rationalize with conjugates
Logarithm	ax = b ⇔ x = loga b
Log laws	product, quotient, power
Change of base	loga x = log x / log a
Equations	match bases, take logs, substitute; check validity

Before the exam, make sure you…

•show each law used
•convert between forms confidently
•spot quadratics in disguise
•reject invalid log solutions
•give exact answers on Paper 1

Exponents and Logarithms on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Exponents and Logarithms.

22 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. x11

2. 1/25; 27/8

3. 50ab6

4. x4/y6

5. 38x+2

6. 2x5/(3y5)

7. 7n ÷ 7n = 1 but also = 7n−n = 70

Practice exercise 2

1. 27

2. 1/16

3. x−3/2

4. x + 2 + 1/x

5. x = 16

6. x7/6

Practice exercise 3

1. 3√3

2. 3√2

3. 4

4. 3 − √5

5. 3 + 2√2

6. (√2 − 1) + (√2 + 1) = 2√2 after rationalizing each

Practice exercise 4

1. −4

2. 2/3

3. x = e3

4. 5

5. 7; 9

6. no power of 2 is negative

Practice exercise 5

1. 3

2. log 10 = 1

3. 2 + 2 log x − ½ log y

4. a − b

5. loga 1 − loga x = 0 − loga x

6. 3/2

Practice exercise 6

1. 2.01

2. log4 x = log2 x / log2 4 = log2 x / 2

3. x2 = x + 12 gives x = 4 (reject −3)

4. 3

Practice exercise 7

1. x = 5/3

2. 2.81

3. x = 2 ln 3 / ln(5/3)

4. x = 0 or x = 1

5. x = ln 2 or ln 3

6. 18x = 100: x = 1.59

Practice exercise 8

1. 9

2. x = 1/9

3. 4

4. x = 9

5. x = 1 or x = e3

Practice exercise 9

1. 14.2 years

2. 102.5 ≈ 316

3. 3.98 × 10−4 mol dm−3

Practice exercise 10

1. 3/2

2. 3x + 2 = 5: x = 1

3. −4

Practice exercise 11

1. x = 0 or 2

2. e2x − ex − 6 = 0: x = ln 3

3. 2y2 − 9y + 4 = 0: y = 4 or 1/2: x = 2 or −1

Practice exercise 12

1. x = 2

2. x2 = 8x − 16: x = 4

3. x + 6 = 4x: x = 2

Practice exercise 13

1. log 2 = 1 − k

2. change of base: (ln b/ln a)(ln a/ln b) = 1

Practice exercise 14

1. k = ln 2/12 = 0.0578

2. ln 1.6/6 = 7.83%

Practice exercise 15

1. x > 1; x = 1; (2, 0)

2. f−1(x) = ½ ln(x − 1), x > 1

Practice exercise 16

1. x = 0 or ln 2

2. x = ln 3/ln 1.5 = 2.71

3. 3

Practice exercise 17

1. x = √11

2. 5x = 5: x = 1

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