DP Maths AI SL 1.5 Exponential and Logarithmic Functions

DP Maths AI SL 1.5 Exponential and Logarithmic Functions - Full notes

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DP Maths AI SL | 1.5 Exponential and Logarithmic Functions

DP MATHEMATICS AI · TOPIC 1: NUMBER AND ALGEBRA

1.5

Exponential and Logarithmic Functions

Full notes for Applications and Interpretation SL

Contents

1	Laws of exponents
2	Negative and fractional exponents
3	Standard form and calculations with large numbers
4	Exponential functions and their graphs
5	The number e
6	Exponential growth models
7	Exponential decay and half-life
8	Introducing logarithms
9	Logarithms with technology
10	Solving exponential equations
11	Logarithmic functions and their graphs
12	Transformed exponential graphs and asymptotes
13	Exponential models of the form a·bˣ + c
14	Fitting exponential models to data
15	Comparing linear and exponential models
16	Solving equations graphically
17	Applications in science and finance
18	Mixed review: worked problems
19	Investigation: radiocarbon dating
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	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 practice exercise.

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 negative and fractional exponents
work with numbers in standard form
sketch exponential functions and identify asymptotes and intercepts
use e and exponential models for growth and decay, including half-life
understand logarithms (base 10 and base e) as inverses of exponentials
solve exponential equations using logarithms or technology
fit exponential models to data and interpret parameters
compare linear and exponential models

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can exponential functions describe growth and decay in the real world? How do logarithms 'undo' exponentials?

How this topic is examined

Exponential functions are among the most important models in the AI course. Expect questions that find a model from information, interpret parameters such as the initial value, growth factor and asymptote, solve for time using logarithms or technology, and comment on the validity of the model.

Show the equation you solve, state the method (logarithms, GDC solver or graph), and interpret every answer in the context of the problem.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

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, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
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
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), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Laws of exponents

Rules for powers.

Think about it

210 is about 1000. Roughly how big is 220? 240? Why do powers grow so quickly?

In an, a is the base and n the exponent. For a positive integer n, an means n factors of a. Counting factors gives the laws:

These laws only apply to powers of the same base (or the same exponent, for products). A common error is to add exponents of different bases.

WORKED EXAMPLE 1 Simplifying (basic)

Simplify (a) x5 × x3; (b) 12y7 ÷ 3y2; (c) (2a3)4.

Solution

(a) x8A1

(b) 4y5A1

WORKED EXAMPLE 2 Combining laws (standard)

Simplify (3x2y)3 × 2xy4.

Solution

27x6y3 × 2xy4M1

= 54x7y7A1

Practice exercise 1

1. Simplify p4 × p9 ÷ p6.

2. Simplify (5m2n3)2.

3. Simplify 18a6b2 ÷ (6a2b).

4. Write 8 × 25 as a single power of 2.

2 Negative and fractional exponents

Reciprocals and roots.

These definitions keep the laws consistent: a3 ÷ a3 = 1 = a0; a0 ÷ a2 = a−2 = 1/a2; (a1/2)2 = a, so a1/2 = √a. A negative exponent means a reciprocal, never a negative number.

WORKED EXAMPLE 3 Evaluating (basic)

Evaluate (a) 4−2; (b) 272/3; (c) (1/16)−1/2.

Solution

(a) 1/16A1

(b) (∛27)2 = 9A1

WORKED EXAMPLE 4 Rewriting for a model (standard)

Write 50/√(t3) in the form atn.

Solution

t3/2 in the denominatorM1

= 50t−3/2A1

Practice exercise 2

1. Evaluate 8−1/3.

2. Evaluate 1252/3.

3. Write 1/(x2√x) as a power of x.

4. Evaluate (9/4)−3/2.

3 Standard form and calculations with large numbers

a × 10k.

Numbers in standard form are written a × 10k with 1 ≤ a < 10 and k an integer. They are essential for scientific and economic data that span many orders of magnitude. The laws of exponents apply: multiply the a values and add the powers of 10.

WORKED EXAMPLE 5 Calculation (standard)

The Sun is 1.496 × 1011 m from Earth, and light travels at 3.00 × 108 m/s. How long does sunlight take to reach Earth?

Solution

t = 1.496 × 1011/(3.00 × 108)M1

= 4.99 × 102 s ≈ 8.3 minutesA1

Practice exercise 3

1. Calculate (4.2 × 105) × (3 × 10−2).

2. A population of 7.9 × 109 people consumes 4.5 × 1012 litres of water per day. Find the average per person.

4 Exponential functions and their graphs

Constant multiplicative change.

An exponential function has the form f(x) = kax (a > 0, a ≠ 1), where the variable is in the exponent. Equal increases in x multiply f(x) by the same factor a. If a > 1 the function grows; if 0 < a < 1 it decays.

Key features of y = kax (k > 0): y-intercept (0, k); horizontal asymptote y = 0; domain ℝ; range y > 0; always increasing (a > 1) or always decreasing (0 < a < 1).

Figure 1. Exponential growth and decay curves with asymptote y = 0.

WORKED EXAMPLE 6 Reading features (standard)

For f(x) = 5(0.8)x, state the y-intercept, the asymptote, and whether f increases or decreases.

Solution

y-intercept (0, 5); asymptote y = 0A1

0 < 0.8 < 1, so f decreasesA1

Practice exercise 4

1. State the range of y = 4(3)x.

2. Sketch y = 2−x, labelling the intercept and asymptote.

3. Which grows faster for large x: x10 or 1.1x?

5 The number e

The natural base.

The number e ≈ 2.718 28 arises from continuous compounding: (1 + 1/n)n → e as n → ∞. Functions of the form y = Aekx are used throughout science because their rate of change is proportional to their size. k > 0 gives growth and k < 0 gives decay.

Any exponential ax can be written as ex ln a, so the two forms are interchangeable.

n	(1 + 1/n)n
1	2
12	2.6130
365	2.7146
10 000	2.7181
1 000 000	2.7183

Practice exercise 5

1. Evaluate e2.5 and e−0.4 to 3 s.f.

2. Write 3x in the form ekx.

6 Exponential growth models

Initial value times growth factor.

An exponential growth model has the form N(t) = N0 × at or N(t) = N0ekt. N0 is the initial value (t = 0), a > 1 the growth factor per unit time (a = 1 + p/100 for p% growth), and k the continuous growth rate. The doubling time is the time for N to double.

WORKED EXAMPLE 7 Population model (standard)

A city of 1.2 million grows by 1.8% per year. Write a model and find the population after 15 years.

Solution

P(t) = 1.2(1.018)t millionA1

P(15) = 1.57 millionA1

WORKED EXAMPLE 8 Doubling time (standard)

An investment grows as V = 5000e0.06t. Find the doubling time.

Solution

e0.06t = 2M1

t = ln 2/0.06 = 11.6 yearsA1

Practice exercise 6

1. Bacteria grow as N = 400e0.35t (t in hours). Find N after 6 hours and the doubling time.

2. A town of 15 000 grows by 4% per year. When does it reach 25 000?

7 Exponential decay and half-life

Shrinking by a constant factor.

Decay models have 0 < a < 1 or k < 0: N(t) = N0(0.85)t or N(t) = N0e−0.2t. The half-life is the time for the quantity to halve. For N0e−kt the half-life is ln 2/k; for N0(1/2)t/T the half-life is T.

Figure 2. Half-life of 5 units: the amount halves every 5 units of time.

WORKED EXAMPLE 9 Radioactive decay (standard)

A 200 g sample has half-life 8 days. Find the mass after 30 days.

Solution

m = 200(0.5)30/8M1

= 14.9 gA1

WORKED EXAMPLE 10 Finding a decay rate (exam level)

A medicine's concentration falls from 60 mg/L to 42 mg/L in 3 hours. Assuming C = 60e−kt, find k and the half-life.

Solution

42 = 60e−3k, k = −(1/3)ln 0.7 = 0.119M1 A1

Half-life = ln 2/0.119 = 5.83 hoursA1

Practice exercise 7

1. Caesium-137 has a half-life of 30 years. What fraction remains after 100 years?

2. A car's value is V = 32 000(0.83)t. Find the 'half-life' of its value.

8 Introducing logarithms

What power?

The logarithm answers 'what power?': loga b is the power to which a must be raised to give b. In AI, the important bases are 10 (written log x) and e (written ln x, the natural logarithm).

So log 1000 = 3, log 0.01 = −2, ln e5 = 5 and ln 1 = 0. Logarithms are only defined for positive numbers. Because logs and exponentials are inverses, 10log x = x and eln x = x.

WORKED EXAMPLE 11 Converting forms (basic)

(a) Write 102.5 = 316.2 in log form. (b) Solve log x = 1.7. (c) Solve ln x = −0.5.

Solution

(a) log 316.2 = 2.5A1

(b) x = 101.7 = 50.1A1

Practice exercise 8

1. Evaluate log 100 000 and log 0.001.

2. Solve log x = 3.2.

3. Solve ln x = 2.

4. Explain why log(−5) is undefined.

9 Logarithms with technology

Evaluate and estimate.

The GDC evaluates log and ln directly. Logarithms to other bases are found with the change of base rule, loga x = ln x/ln a, or with a loga template. Logarithms compress large ranges: the numbers 1 to 1 000 000 have base-10 logarithms 0 to 6.

WORKED EXAMPLE 12 Order of magnitude (standard)

Use logarithms to find the number of digits in 2100.

Solution

log 2100 = 100 log 2 = 30.1M1

So 2100 ≈ 1030.1: 31 digitsA1

Practice exercise 9

1. Evaluate log3 50.

2. How many digits does 750 have?

10 Solving exponential equations

Take logs or use technology.

To solve ax = b, take logarithms of both sides: x = log b/log a (or ln b/ln a). For models such as 500(1.04)t = 800, first isolate the power. On Paper 2, a graph or the solver is equally acceptable, provided you show the equation you solved.

WORKED EXAMPLE 13 Solving (standard)

Solve 3 × 2x = 50.

Solution

2x = 50/3M1

x = ln(50/3)/ln 2 = 4.06A1

WORKED EXAMPLE 14 Model equation (exam level)

Solve 1200e−0.15t + 300 = 700.

Solution

e−0.15t = 1/3M1

t = ln 3/0.15 = 7.32A1

Practice exercise 10

1. Solve 5x = 200.

2. Solve 40(0.9)t = 15.

3. Solve 2e0.3x = 18.

11 Logarithmic functions and their graphs

Inverses of exponentials.

The graph of y = log x (or ln x) is the reflection of y = 10x (or ex) in y = x. Features of y = ln x: x-intercept (1, 0); vertical asymptote x = 0; domain x > 0; range ℝ; increasing, but ever more slowly.

Figure 3. y = ln x is the reflection of y = ex in y = x.

Practice exercise 11

1. State the domain and asymptote of y = ln(x − 3).

2. Find the x-intercept of y = 2 + log x.

12 Transformed exponential graphs and asymptotes

Shifts and stretches.

For y = kax + c, the horizontal asymptote is y = c, the y-intercept is k + c, and the range is y > c (if k > 0) or y < c (if k < 0). Recognizing the asymptote is essential for interpreting models: it is the long-term value.

Figure 4. A limited-growth model approaching y = 20.

WORKED EXAMPLE 15 Features (standard)

For f(x) = 8 − 3e−0.5x, find the asymptote, intercept and range for x ≥ 0.

Solution

Asymptote y = 8A1

f(0) = 5A1

Range 5 ≤ f(x) < 8A1

Practice exercise 12

1. State the asymptote and y-intercept of y = 4(2)x − 6.

2. Find the range of y = 10 + 5e−x for x ≥ 0.

13 Exponential models of the form a·bˣ + c

Cooling, learning and limited growth.

Models of the form T(t) = c + a bt (with 0 < b < 1) approach the value c. Newton's law of cooling gives T = Troom + (T0 − Troom)e−kt. Learning curves and the absorption of medicine follow similar shapes. Interpret c as the long-term value and c + a as the initial value.

WORKED EXAMPLE 16 Cooling coffee (exam level)

T(t) = 21 + 64(0.94)t °C, t in minutes. Find the initial temperature, the temperature after 15 minutes, the long-term temperature and when T = 40.

Solution

T(0) = 85 °C; T(15) = 46.3 °CA1 A1

Long-term 21 °C (room temperature)A1

64(0.94)t = 19: t = 19.6 minutesM1 A1

Practice exercise 13

1. A learning curve is P(n) = 95 − 70(0.8)n, where P is the percentage correct after n practice sessions. Find P(5) and the number of sessions needed to exceed 90%.

2. Find k if T = 20 + 60e−kt and T = 50 when t = 8.

14 Fitting exponential models to data

Exponential regression.

When data appear to grow or decay by a roughly constant factor, a GDC can fit an exponential regression model y = abx. The coefficient of determination R2 (or r for the linearized data) indicates the quality of fit. Always plot the data and the model together and comment on the fit.

Year after 2015, x	0	2	4	6	8
Subscribers (thousands), y	5.2	8.1	12.9	20.4	31.8

WORKED EXAMPLE 17 Exponential regression (exam level)

Fit an exponential model to the subscriber data and predict x = 10.

Solution

GDC: y = 5.18(1.255)xA1 A1

Growth about 25.5% per yearR1

x = 10: 50.3 thousandA1

Figure 5. Data and the fitted exponential model.

Practice exercise 14

1. Explain why exponential regression is preferred to linear regression for the subscriber data.

2. Suggest one reason why the model may fail for large x.

15 Comparing linear and exponential models

Differences versus ratios.

For equally spaced x values: constant differences in y suggest a linear model; constant ratios suggest an exponential model. Exponential growth eventually exceeds any linear growth, however small the growth rate.

WORKED EXAMPLE 18 Which model? (standard)

Values at x = 0, 1, 2, 3 are 40, 52, 67.6, 87.9. Decide which model fits.

Solution

Differences 12, 15.6, 20.3: not constantM1

Ratios 1.3, 1.3, 1.3: exponential, y = 40(1.3)xA1

Practice exercise 15

1. Values 15, 22, 29, 36: which model?

2. When does 100(1.05)x first exceed 200 + 10x?

16 Solving equations graphically

Intersections on the GDC.

Equations mixing exponentials with polynomials, such as 2x = x2 or ex = 4 − x, cannot be solved algebraically. Graph both sides and find the intersections; report all solutions in the required domain.

Figure 6. 2x = x2 has three solutions.

WORKED EXAMPLE 19 Graphical solution (standard)

Solve 2x = x2.

Solution

Intersections from the graphM1

x = −0.767, x = 2, x = 4A2

Practice exercise 16

1. Solve ex = 4 − x.

2. Solve ln x = 0.5x − 1.

17 Applications in science and finance

Exponentials everywhere.

Continuous compounding: A = Pert gives slightly more than any finite compounding. Atmospheric pressure decreases roughly exponentially with altitude. Light intensity decreases exponentially through water. Cooling, charging capacitors, drug elimination and the spread of early epidemics are all modelled with exponentials.

WORKED EXAMPLE 20 Continuous compounding (standard)

$8000 is invested at 3.5% compounded continuously. Find the value after 10 years.

Solution

8000e0.35M1

= $11 352.54A1

WORKED EXAMPLE 21 Atmospheric pressure (exam level)

P = 101.3e−0.12h kPa at height h km. Find the pressure at 8.8 km and the height at which it is half its sea-level value.

Solution

P(8.8) = 35.2 kPaA1

e−0.12h = 0.5: h = 5.78 kmM1 A1

Practice exercise 17

1. Light intensity falls by 8% per metre of water depth. At what depth is it 10% of the surface value?

2. Compare $1000 at 5% compounded annually and continuously for 20 years.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 22 Mixed problem 1 (exam level)

A model V(t) = Aekt has V(0) = 50 and V(4) = 80. Find A, k and V(10).

Solution

A = 50A1

e4k = 1.6, k = 0.1175M1 A1

V(10) = 161.9A1

WORKED EXAMPLE 23 Mixed problem 2 (exam level)

Two populations: P = 2000(1.06)t and Q = 3500(1.02)t. When are they equal?

Solution

(1.06/1.02)t = 1.75M1

t = ln 1.75/ln(1.06/1.02) = 14.5 yearsA1

Practice exercise 18

1. Solve 6 − 4(0.5)x = 5.

2. A 10 kg sample decays to 7 kg in 5 years. Find the half-life.

19 Investigation: radiocarbon dating

A Paper 3 style exploration.

Living organisms contain carbon-14 at a constant level. After death it decays with a half-life of about 5730 years.

1. Show that the fraction remaining after t years is (1/2)t/5730. 2. Write the model in the form e−kt and find k. 3. A bone contains 35% of the original carbon-14. Estimate its age. 4. Explain why the method becomes unreliable for very old samples (over about 50 000 years). 5. A measurement uncertainty of ±1% in the fraction remaining is typical. Find the corresponding range of ages for the bone in part 3.

Answers. 2. k = ln 2/5730 = 1.21 × 10−4. 3. t = ln(0.35)/(−k) = 8680 years. 4. Very little C-14 remains (under 0.25%), so small measurement errors dominate. 5. 0.34 to 0.36 gives about 8440 to 8920 years.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

The value of a painting is modelled by V(t) = 4000(1.07)t, t years after purchase.

(a) Write down the purchase price. [1]

(b) Interpret the number 1.07. [1]

(d) Find when the value first exceeds $20 000. [3]

Markscheme

Part	Answer	Marks
(a)	$4000	A1
(b)	the value increases by 7% per year	A1
(c)	4000(1.07)12	M1
	$9008.77	A1
(d)	(1.07)t > 5	M1
	t > 23.8	A1
	after 24 years	A1

Examiner's comments. Interpretation must refer to the context.

Question 2 (Paper 2 style) (about 15 minutes)

Water in a kettle cools according to T(t) = 22 + 78e−kt, t in minutes. After 10 minutes T = 60 °C.

(a) Write down the initial temperature. [1]

(b) Find k. [3]

(d) Write down the long-term temperature and interpret it. [2]

Markscheme

Part	Answer	Marks
(a)	100 °C	A1
(b)	38 = 78e−10k	M1
	k = −(1/10)ln(38/78)	A1
	0.0719	A1
(c)	22 + 78e−30k	M1
	31.0 °C	A1
(d)	22 °C	A1
	room temperature	R1

Examiner's comments. Keep k unrounded for (c).

Question 3 (Paper 2 style) (about 10 minutes)

The table gives a country's electric vehicle sales (thousands): year x = 0, 1, 2, 3, 4; sales 14, 19, 27, 37, 52.

(a) Use technology to find an exponential regression model y = abx. [2]

(b) Interpret b. [1]

Markscheme

Part	Answer	Marks
(a)	y = 13.9(1.39)x	A2
(b)	sales grow about 39% per year	R1
(c)	y(7) ≈ 139 thousand	A1
	extrapolation	R1
	growth may slow as the market matures	R1

Examiner's comments. Reasonable comments are credited.

21 Common misconceptions and links

Students often think…	Actually…
a−n is negative	a−n = 1/an
the asymptote of kax + c is y = 0	y = c
log(x + y) = log x + log y	log(xy) = log x + log y
exponential models are valid forever	consider limits and context

How 1.5 connects to the rest of the course

Topic	Connection
1.3 Geometric sequences	discrete version
1.9 Logarithms (HL)	laws and log scales
2.9 Transformations	asymptotes
5.14 Differential equations	dy/dt = ky

22 Check your understanding

Section A: multiple choice

1. 2−3 =

A. 1/8 B. −8 C. −6 D. 8

2. The asymptote of y = 3(0.5)x + 4 is

A. y = 0 B. y = 3 C. y = 4 D. x = 4

3. ln e2 =

A. e2 B. 2 C. 1 D. 7.39

4. Half-life for N = N0e−0.1t is

A. 0.1 B. 10 C. 5 D. 6.93

5. Constant ratios in data suggest

A. a linear model B. an exponential model C. a quadratic model D. no model

Section B: structured

6. Solve 250(1.12)t = 1000. [2]

7. A sample decays from 80 g to 50 g in 6 hours. Find the half-life. [3]

Answers

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

6. t = ln 4/ln 1.12 = 12.2.

7. k = ln 1.6/6 = 0.0783; half-life 8.85 hours.

23 Summary and topic map

Idea	Remember
Exponent laws	add, subtract, multiply exponents
Negative/fractional	a−n = 1/an; a1/n = nth root
Exponential model	N0at or N0ekt
Asymptote	y = c for kax + c
Half-life	ln 2/k
Logs	log = base 10, ln = base e; inverses of exponentials
Solving	take logs or use technology

Before the exam, make sure you…

•interpret parameters in context
•identify the asymptote
•keep constants unrounded
•comment on model validity

Exponential and Logarithmic Functions 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 7. Topic map for Exponential and Logarithmic Functions.

24 Answers to practice exercises

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

Practice exercise 1

1. p7

2. 25m4n6

3. 3a4b

4. 28

Practice exercise 2

1. 1/2

2. 25

3. x−5/2

4. 8/27

Practice exercise 3

1. 1.26 × 104

2. 570 litres

Practice exercise 4

1. y > 0

2. decay through (0, 1), asymptote y = 0

3. 1.1x eventually exceeds any power of x

Practice exercise 5

1. 12.2; 0.670

2. e1.10x (k = ln 3)

Practice exercise 6

1. 3266; 1.98 hours

2. 13.0 years

Practice exercise 7

1. (0.5)100/30 = 0.0992

2. ln 0.5/ln 0.83 = 3.72 years

Practice exercise 8

1. 5; −3

2. 1585

3. 7.39

4. no power of 10 is negative

Practice exercise 9

1. 3.56

2. 50 log 7 = 42.3: 43 digits

Practice exercise 10

1. 3.29

2. 9.31

3. 7.32

Practice exercise 11

1. x > 3; x = 3

2. x = 0.01

Practice exercise 12

1. y = −6; (0, −2)

2. 10 < y ≤ 15

Practice exercise 13

1. 72.1%; n > 11.8, so 12 sessions

2. k = ln 2/8 = 0.0866

Practice exercise 14

1. ratios are roughly constant; differences increase

2. the market saturates; growth slows

Practice exercise 15

1. linear, d = 7

2. x = 34.9, so x = 35

Practice exercise 16

1. 1.07

2. 0.464 and 5.36

Practice exercise 17

1. 0.92d = 0.1: 27.6 m

2. $2653.30 vs $2718.28

Practice exercise 18

1. x = 2

2. 9.72 years

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