Exponentials and logarithms. Attnom Academy

M5.2.1 Chapter 1

Exponential functions

Can a number such as 2ˣ ever be zero or negative, and what does its graph look like between the whole numbers?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Yes. It becomes negative when x is negative.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes, but only when x is 0.
You picked B. Read on. At the end, the chapter comes back to your guess.
C No. It is above zero for every x.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sketch y = 3ˣ and y = (1/2)ˣ on one grid
Explain why aˣ is always positive

You have met 2ˣ for whole numbers x, such as 2⁵ = 32 and 2⁻² = 1/4. Now x may be any real number. The rule f(x) = aˣ is an exponential function. The number a is its base, and x is the power.

The base must be above 0 and not equal to 1. Take a = −2. Then (−2)² = 4 and (−2)³ = −8, but the power 0.5 would need the square root of −2, which is not a real number. Take a = 1. Then 1ˣ = 1 for every x, a flat line and not a new function. So we always write a > 0 and a ≠ 1.

Values of 3ˣ and 0.5ˣ. Each step right multiplies the first column by 3 and the second by 0.5.

Between the whole numbers the rule keeps the laws of powers. A power of 1/2 means a square root, so 2 to the power 0.5 is the square root of 2. A power of 3/2 means a square root, then a cube. For a power such as √2, which has endless digits, we use closer and closer decimals. The values settle on one number, so aˣ has a value for every real x.

As x gets closer to √2, which is about 1.41421, the value of 2ˣ settles at 2.6651 to 4 decimal places. So 2 to the power √2 is about 2.665.

Worked example

y = 4ˣ. Find y when x = 1.5, and when x = −0.5.

A power of 1.5 is 3/2. Take the square root of 4 first: 4 to the power 0.5 is 2.
Then cube it: 4 to the power 1.5 is 2³ = 8.
A negative power means one over the positive power: 4 to the power −0.5 is 1 ÷ 4 to the power 0.5.
So 4 to the power −0.5 is 1 ÷ 2 = 0.5.

Answer: When x = 1.5, y = 8. When x = −0.5, y = 0.5.

Draw it first

The graph of y = 3ˣ

1 Plot the points from the table. On the left they sit very close to the x axis.
2 Join them with one smooth curve. It passes through (0, 1) and rises faster and faster.

The growth curve y = 3ˣ and the decay curve y = 0.5ˣ, which is the same as y = (1/2)ˣ. Both pass through A at (0, 1).

Read the graph. With a above 1 the curve is a growth curve: it rises as x rises. With a between 0 and 1 it is a decay curve: it falls as x rises. In both cases the curve crosses the y axis at (0, 1), because a⁰ = 1. It also gets as close as you like to the x axis on one side, without touching it. A line that a curve approaches but never reaches is an asymptote. Here the asymptote is the x axis, the line y = 0.

The curves y = 2ˣ and y = 0.5ˣ are mirror images in the y axis. P is (−1, 2) and Q is (1, 2).

The two curves in the last picture are mirror images because 0.5ˣ = (2⁻¹)ˣ = 2⁻ˣ. Swapping x for −x reflects a graph in the y axis. In general y = (1/a)ˣ is the reflection of y = aˣ in the y axis. For x above 0 a bigger base gives a steeper curve. For x below 0 the order is reversed, and 3ˣ lies below 2ˣ.

Now the reason aˣ is always positive. For a whole x above 0, aˣ is a product of positive numbers. For x below 0, aˣ is one over a positive number. A fractional power is a positive root, by our definition. For the rest, here is one argument that works for every x. First, aˣ is the square of a to the power x/2, and a square is never negative. Second, aˣ × a⁻ˣ = a⁰ = 1. A product equal to 1 has no zero factor, so aˣ is not 0. A number that is not negative and not zero is above zero. So aˣ > 0 for all real x.

Worked example

Sketch y = 2ˣ − 4. Give where it crosses each axis, its asymptote and its range.

The graph of y = 2ˣ − 4 is the graph of y = 2ˣ moved down 4 units.
Cross the y axis: put x = 0. Then y = 2⁰ − 4 = 1 − 4 = −3. The point is (0, −3).
Cross the x axis: put y = 0. Then 2ˣ − 4 = 0, so 2ˣ = 4. This gives x = 2.
On the far left 2ˣ is close to 0, so y is close to −4. The asymptote is the line y = −4.
Since 2ˣ > 0, y = 2ˣ − 4 is above −4. The range is y > −4.

Answer: y-intercept (0, −3), x-intercept (2, 0), asymptote y = −4, range y > −4.

The graph of y = 2ˣ − 4 with its asymptote, the dashed line y = −4.

See it three ways

With things

A dose of 8 mg of a medicine is in the blood. Each hour half of what is there leaves. After 1 hour 4 mg is left, after 2 hours 2 mg, after 3 hours 1 mg.

As a picture

In words

The dose halves every hour, so it falls fast at first and then slowly. It gets close to 0 but is never exactly 0.

In symbols

D = 8 × 0.5ˣ, where x is the time in hours. So D = 8 when x = 0, D = 4 when x = 1, D = 2 when x = 2 and D = 1 when x = 3.

Try it

Watch 2ˣ fill the gaps

You need: a scientific calculator, paper and a pencil.

Work out 2¹, then 2 to the power 1.5, then 2² on the calculator. Use the power key.
Now work out 2 to the power 1.9 and 2 to the power 2.1.
Write the five results in order of x.

What you will see: The values climb smoothly: 2, 2.83, 3.73, 4, 4.29. There are no gaps and no jumps. The values for 1.9 and 2.1 sit just either side of 4.

If you cannot try it, this is what happens.

A wrong idea many people have

A negative power gives a negative number, so 2⁻³ = −8.

Why it feels right: The minus sign in the power looks like it should turn the answer negative.

The test that breaks it: Use 2⁻¹ = 1 ÷ 2 = 0.5, 2⁻² = 0.25 and 2⁻³ = 0.125. Each step down divides by 2, and dividing a positive number by 2 stays positive.

What is true: A negative power means one over the positive power. It is small, but always above zero. Catch the slip by asking: is this number a product, a root or a reciprocal of positive numbers?

A wrong idea many people have

The graph of a decay curve crosses the x axis somewhere on the right.

Why it feels right: The curve falls and falls, so it seems it must reach 0.

The test that breaks it: Take y = 0.5ˣ. At x = 10 it is about 0.00098, at x = 20 about 0.00000095. Each value is half of one still above 0.

What is true: The x axis is an asymptote: the curve gets as close as you like but never touches it. The range of y = aˣ is y > 0, and the equation aˣ = 0 has no solution.

A wrong idea many people have

y = 2ˣ is the same kind of curve as y = x².

Why it feels right: Both climb quickly, and both have a 2 and an x.

The test that breaks it: Compare x = 3 and x = 10. For x² you get 9 and 100. For 2ˣ you get 8 and 1 024. Now try x = −3: x² gives 9, but 2ˣ gives 0.125.

What is true: In x² the letter is the base and the power is fixed. In 2ˣ the letter is the power and the base is fixed. The curve y = x² is symmetric and touches the x axis at 0. The curve y = 2ˣ is not symmetric and stays above the x axis.

Key words

exponential function: a rule f(x) = aˣ with a above 0 and a ≠ 1, where x is the power.
base: the number a in aˣ, which is multiplied by itself.
growth: the shape of y = aˣ when a is above 1. The values rise as x rises.
decay: the shape of y = aˣ when a is between 0 and 1. The values fall as x rises.
asymptote: a line that a curve gets closer to but never reaches.

Back to your guess

The answer is C: No. It is above zero for every x.

A negative power does not make a negative number. 2⁻³ means 1 ÷ 2³ = 1/8, which is small but above zero. And 2⁰ = 1. So 2ˣ is above zero for every x, and the graph never reaches the x axis.

Keep it

Copy this into your exercise book.

y = aˣ, a > 0, a ≠ 1: passes through (0, 1), always above the x axis. a > 1 grows, 0 < a < 1 decays. The x axis is an asymptote.

Draw: A rising curve and a falling curve crossing at (0, 1), with the x axis dashed.

M5.2.2 Chapter 2

Logarithms

We can solve 2ˣ = 32 by counting powers of 2. How do we write, and find, the answer to 2ˣ = 20?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A x = 16, because 32 ÷ 2 = 16.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x = 5, because 2 × 2 × 2 × 2 × 2 = 32.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x = 64, because 32 × 2 = 64.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Change 2⁵ = 32 into log₂ 32 = 5
Work out log₃ 81 without a calculator

In the last chapter you found x from 2ˣ − 8 = 0 by trying powers: 2³ = 8. That works when the answer is a whole number. For 2ˣ = 20 no whole number works, because 2⁴ = 16 and 2⁵ = 32. The answer lies between 4 and 5. We need a name for it.

The logarithm of a number to a base is the power you must raise the base to, to get the number. In symbols, for a > 0, a ≠ 1 and b > 0: the log of b to the base a is x means aˣ = b. With base 2 we write log₂ 32 = 5, and say "log to the base 2 of 32 is 5". The name log is short for logarithm.

The same fact written two ways. The base stays the base. The power becomes the answer. The number goes inside the log.

The two forms say exactly the same thing. The first, 2⁵ = 32, is the index form. The second, log₂ 32 = 5, is the log form. To change one into the other, find the base, the power and the number, and put each in its new place as in the picture.

The switch for 2⁷ = 128.

Worked example

Write 2⁷ = 128 as a logarithm. Then write log₇ 49 = 2 as a power.

In 2⁷ = 128 the base is 2, the power is 7 and the number is 128.
The log of the number to the base is the power: log₂ 128 = 7.
In log₇ 49 = 2 the base is 7, the number is 49 and the answer, 2, is the power.
So the index form is 7² = 49.

Answer: log₂ 128 = 7 and 7² = 49.

Now work out a logarithm. To find log₃ 81, ask: 3 to which power gives 81? Count the powers of 3 upwards until you reach 81. If the number is a fraction, use negative powers. The next example does both.

Worked example

Work out log₃ 81 and log₂ (1/8) without a calculator.

log₃ 81 asks: 3 to which power gives 81?
Count the powers of 3: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81.
So log₃ 81 = 4.
log₂ (1/8) asks: 2 to which power gives 1/8?
A negative power gives a fraction. Since 2³ = 8, we have 2⁻³ = 1/8.
So log₂ (1/8) = −3.

Answer: log₃ 81 = 4 and log₂ (1/8) = −3.

Some answers are given by the definition, with no counting. Since a⁰ = 1, the log of 1 to any base is 0. Since a¹ = a, the log of a to the base a is 1. More generally the log of aⁿ to the base a is n, because n is the power that gives aⁿ. And a to the power (the log of b to the base a) is b, because that log is the power that gives b.

There are limits. The number inside a log must be above 0. The reason is in chapter 1: aˣ is above 0 for every x, so no power of a gives 0 or a negative number. So log₂ 0 and log₂ (−4) do not exist. The base must be above 0 and not 1. If a = 1 then 1ˣ = 1 for every x, so log₁ 5 has no answer and log₁ 1 has too many.

Values of log₂ x. Each time x doubles, log₂ x goes up by 1.

Draw it first

The graph of y = log₂ x

1 Plot the points (x, log₂ x) from the table.
2 Join them with one smooth curve. It is steep near the y axis and then flattens, but it never stops rising.

The graph of y = log₂ x is the mirror image of y = 2ˣ in the line y = x. This is because log₂ is the inverse function of 2ˣ: it undoes it. If 2ˣ takes 3 to 8, then log₂ takes 8 back to 3. So each point (p, q) on y = 2ˣ gives a point (q, p) on y = log₂ x. The y axis is the asymptote of y = log₂ x, the way the x axis is for y = 2ˣ.

y = 2ˣ and y = log₂ x are mirror images in the line y = x. A is (3, 8) and B is (8, 3).

Calculators have a log key for base 10. In this book log with no base written means base 10. Some books use log for the natural logarithm, which you meet in chapter 5, so check your own paper. By the definition, log 10 = 1, log 100 = 2 and log 1 000 = 3. A value such as log 50 is not a whole number, and the calculator gives it as a decimal.

Logarithms turn multiplying into counting steps. In chemistry the pH of a solution is pH = −log₁₀ c, where c is the concentration of hydrogen ions in mol/dm³. A solution with c = 0.001 has pH 3. A solution with c = 0.00001 has pH 5, and its concentration is 100 times smaller. Each step of 1 on the pH scale is a factor of 10.

Worked example

Find x in each case. (a) The log to the base 5 of x is 3. (b) The log to the base x of 64 is 3. (c) log₂ (3x + 1) = 4.

(a) Change to index form: x = 5³ = 125.
(b) Change to index form: x³ = 64. The cube root of 64 is 4, so x = 4. A base must be above 0, and 4 is.
(c) Change to index form: 3x + 1 = 2⁴ = 16.
Take 1 from both sides: 3x = 15. Divide by 3: x = 5.
Check (c): 3 × 5 + 1 = 16 is above 0, and log₂ 16 = 4. So x = 5 works.

Answer: (a) x = 125, (b) x = 4, (c) x = 5.

See it three ways

With things

A pile of seeds starts at 1 and doubles every day: 1, 2, 4, 8, 16, 32. To reach 32 it takes 5 doublings.

As a picture

In words

The logarithm to the base 2 of the number of seeds counts how many doublings it took to get there.

In symbols

Since 2⁵ = 32, log₂ 32 = 5. Since 2³ = 8, log₂ 8 = 3.

Try it

Find logs on a calculator

You need: a scientific calculator, paper and a pencil.

Press the log key, then 10, to find log 10.
Find log 100, then log 1000.
Find log 50 and log 5000. Write each result.

What you will see: The first three results are 1, 2 and 3. Each time the number is 10 times bigger, the log goes up by 1. The log of 50 is between 1 and 2, and the log of 5 000 is between 3 and 4.

If you cannot try it, this is what happens.

A wrong idea many people have

log₃ 81 = 81 ÷ 3 = 27.

Why it feels right: The word "log" and the 3 suggest a division by 3.

The test that breaks it: If log₃ 81 were 27, then 3 to the power 27 would be 81. But 3² = 9 and 3³ = 27 already, and 3²⁷ is a number with 13 digits. The correct power is the one that gives 81: 3⁴ = 81.

What is true: A log is a power, not a quotient. To find the log of b to the base a, ask: a to which power gives b? Catch the slip by changing it to index form and checking: 3²⁷ is not 81.

A wrong idea many people have

log₂ (−4) = −2, because 2² = 4.

Why it feels right: The number 4 appears, and a minus sign seems to move out to the answer.

The test that breaks it: If log₂ (−4) = −2 then 2⁻² would be −4. But 2⁻² = 1/4. No power of 2 gives −4, since every power of 2 is above 0.

What is true: A log exists only for a number above 0. Before you work out a log, check that the number inside is above 0. A calculator shows an error for log (−4).

A wrong idea many people have

log₂ 8 = 4, because 2 × 4 = 8.

Why it feels right: You are looking for a number that goes with 2 to make 8, and 2 × 4 is a familiar fact.

The test that breaks it: Change to index form. If log₂ 8 = 4 then 2⁴ = 8. But 2⁴ = 16. The power that gives 8 is 3, since 2 × 2 × 2 = 8.

What is true: The answer to a log is a power, so count how many times the base is multiplied, not what it is multiplied by. Catch the slip by changing to index form and checking.

Key words

logarithm: the power a base must be raised to, to give a number. The log of b to the base a is x when aˣ = b.
index form: a statement such as 2⁵ = 32, with a power.
log form: the same statement written with a log, log₂ 32 = 5.
inverse function: a function that undoes another. The log to the base a undoes aˣ.

Back to your guess

The answer is B: x = 5, because 2 × 2 × 2 × 2 × 2 = 32.

The power of 2 that gives 32 is 5, since 2⁵ = 32. We give this fact a name: log₂ 32 = 5, said "log to the base 2 of 32 is 5". For 2ˣ = 20 the power is between 4 and 5, and its name is log₂ 20. This chapter shows how to write, read and use such powers.

Keep it

Copy this into your exercise book.

The log of b to the base a is x means aˣ = b (a > 0, a ≠ 1, b > 0). The log of 1 is 0. The log of the base is 1. The log undoes aˣ.

Draw: The index line over the log line with three arrows: base, power, number.

M5.2.3 Chapter 3

Laws of logarithms

Logarithms are powers, so they should follow the laws of powers. What do they let us do with a product, a quotient and a power?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A log 10 + log 5 = log 50.
You picked A. Read on. At the end, the chapter comes back to your guess.
B log 10 + log 5 = log 15.
You picked B. Read on. At the end, the chapter comes back to your guess.
C log 10 + log 5 = (log 10) × (log 5).
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use the laws for the log of a product, quotient and power
Write 2 log 3 + log 4 as one logarithm

The laws of indices say a³ × a⁴ = a⁷: to multiply powers of one base, add the powers. A logarithm is a power. So adding logarithms should match multiplying numbers. That is the first law, the product law, and the picture shows it with lengths.

Lengths drawn to scale, in base 10. The bar for log 2 and the bar for log 3, end to end, are as long as the bar for log 6.

Here are the three laws. They hold for logs to any one base a, with a > 0 and a ≠ 1, as long as every number inside a log is above 0. The base must be the same throughout.

The product law, the quotient law and the power law, for x above 0 and y above 0. The power n can be any real number.

Worked example

Prove the product law: for x > 0 and y > 0, log (xy) = log x + log y, with every log to the base a.

Let p be the log of x to the base a, and q the log of y. By the definition, x = a to the power p and y = a to the power q.
Multiply: xy = (a to the power p) × (a to the power q).
The law of indices says to add the powers: xy = a to the power (p + q).
Read that as a logarithm: the power of a that gives xy is p + q. So the log of xy is p + q.
Put back p and q: log (xy) = log x + log y.

Answer: log (xy) = log x + log y.

The other two laws come the same way. For the quotient law, x ÷ y is (a to the power p) ÷ (a to the power q), which is a to the power (p − q). So the log of x ÷ y is p − q. For the power law, xⁿ is (a to the power p) raised to n, which is a to the power np. So the log of xⁿ is np. Two special cases are worth knowing: n = −1 gives log (1/x) = −log x, and n = 1/2 gives log √x = 1/2 log x.

Take the bar for log 3 away from the bar for log 12 and the bar for log 4 is left. That is the quotient law: log 12 − log 3 = log 4.

Worked example

Write 2 log 3 + log 4 as one logarithm.

Use the power law on the first term: 2 log 3 = log (3²) = log 9.
The sum is now log 9 + log 4.
Use the product law and multiply the numbers inside: log 9 + log 4 = log (9 × 4) = log 36.
So 2 log 3 + log 4 can be written as one logarithm, log 36.

Answer: 2 log 3 + log 4 = log 36.

Worked example

Given that log 2 = a and log 3 = b, with every log to base 10, write each of these in terms of a and b. (i) log 72. (ii) log 1.5. (iii) log 5.

(i) 72 = 2³ × 3². So log 72 = log 2³ + log 3² = 3 log 2 + 2 log 3 = 3a + 2b.
(ii) 1.5 = 3 ÷ 2. So log 1.5 = log 3 − log 2 = b − a.
For (iii) we need 5 in terms of 2. Use 5 = 10 ÷ 2.
In base 10 the log of 10 is 1. Since 10¹ = 10, the log of 10 is 1.
(iii) log 5 = log 10 − log 2 = 1 − a.

Answer: (i) 3a + 2b, (ii) b − a, (iii) 1 − a.

Worked example

Show that log₂ 48 − log₂ 3 = 4.

Both logs are to the base 2, so the quotient law applies.
log₂ 48 − log₂ 3 = log₂ (48 ÷ 3).
Work out the division: 48 ÷ 3 = 16.
Since 2⁴ = 16, log₂ 16 = 4.
So log₂ 48 − log₂ 3 = 4, as required.

Answer: log₂ 48 − log₂ 3 = log₂ 16 = 4.

There is one more rule, for a calculator that has only a base 10 key and a natural log key. The change of base formula lets you change the base of any logarithm. Here is the proof, then an example.

Worked example

Prove that the log of x to the base a is log x ÷ log a. The logs on the right are to any one base.

Let y be the log of x to the base a. By the definition, a to the power y is x.
Take the log of both sides, to any base: log (a to the power y) = log x.
The power law moves y to the front: y log a = log x.
Divide both sides by log a. This is allowed, because a ≠ 1 means log a is not 0: y = log x ÷ log a.
So the log of x to the base a is log x ÷ log a.

Answer: The log of x to the base a is log x ÷ log a.

Worked example

Find log₅ 40 to 3 significant figures.

The calculator log key is base 10, so change the base: log₅ 40 = log 40 ÷ log 5.
log 40 = 1.60206 and log 5 = 0.69897, each to 5 decimal places.
Divide: 1.60206 ÷ 0.69897 = 2.292.
To 3 significant figures, log₅ 40 = 2.29.
Check: 5² = 25 and 5³ = 125, and 40 is between them. So the answer must be between 2 and 3, and it is.

Answer: 2.29

See it three ways

With things

Two sticks are cut. One is log 4 units long, which is 0.602 units. The other is log 5 units long, which is 0.699 units. Laid end to end they are 1.301 units long, which is log 20.

As a picture

In words

Adding the lengths of two logs gives the length of the log of the product. Multiplying numbers has become adding lengths.

In symbols

log 4 + log 5 = log (4 × 5) = log 20, and 0.602 + 0.699 = 1.301.

Try it

Test the product law

You need: a scientific calculator, paper and a pencil.

Find log 8 and log 5. Write each to 3 decimal places.
Add the two results.
Now find log 40, which is 8 × 5, to 3 decimal places. Compare it with the sum.

What you will see: The sum of log 8 and log 5 and the value of log 40 are the same, 1.602. Adding logs gives the log of the product. Now try log 8 × log 5, and you get a different number.

If you cannot try it, this is what happens.

A wrong idea many people have

log (a + b) = log a + log b.

Why it feels right: It looks like the way brackets spread in 2(a + b) = 2a + 2b. The log seems to be handed out to each term.

The test that breaks it: Put a = 10 and b = 10 in base 10. Then log (10 + 10) = log 20 = 1.301. But log 10 + log 10 = 1 + 1 = 2. The two sides are not equal.

What is true: A log is not a number that multiplies a bracket. It is a function, and log (a + b) cannot be split. The law adds logs only for a product: log a + log b = log (ab). Catch the slip by asking what the sign inside the bracket is. Only a product or a quotient inside lets a law work.

A wrong idea many people have

log x ÷ log y = log (x ÷ y), or log x − log y = log x ÷ log y.

Why it feels right: The quotient law has a division in it, so it is easy to put the division in the wrong place.

The test that breaks it: In base 10 take x = 100 and y = 10. Then log x ÷ log y = 2 ÷ 1 = 2. But log (x ÷ y) = log 10 = 1, and log x − log y = 2 − 1 = 1.

What is true: The difference of two logs is the log of the quotient: log x − log y = log (x ÷ y). The quotient of two logs, log x ÷ log y, has no simpler form. It appears only in the change of base formula.

A wrong idea many people have

(log x)² = 2 log x, and log (xy) = (log x)(log y).

Why it feels right: The power law moves a power to the front, and the product law has two logs in it. The pieces look interchangeable.

The test that breaks it: In base 10 take x = 10. Then (log x)² = 1² = 1, but 2 log x = 2. Also take x = 10 and y = 100: log (xy) = log 1 000 = 3, but (log x)(log y) = 1 × 2 = 2.

What is true: The power law is about a power inside the log: log (x²) = 2 log x. The square of a log, (log x)², is a different number. And the log of a product is the sum of the logs, not their product.

Key words

product law: log x + log y = log (xy), for logs to the same base and x, y above 0.
quotient law: log x − log y = log (x ÷ y).
power law: n log x = log (xⁿ), where n is any real number.
change of base formula: the log of x to the base a is log x ÷ log a, with both logs to any one base.

Back to your guess

The answer is A: log 10 + log 5 = log 50.

Adding logs multiplies the numbers inside. In base 10, log 10 = 1 and log 5 = 0.699, so the sum is 1.699. And log 50 = 1.699, because 50 = 10 × 5. The sum is not log 15. This chapter shows why, and what else follows.

Keep it

Copy this into your exercise book.

log xy = log x + log y. log (x/y) = log x − log y. log (xⁿ) = n log x. Same base. No law for log (x + y). Change of base: the log of x to base a is log x ÷ log a.

Draw: Two bars end to end, log x and log y, and one bar under them for log xy.

M5.2.4 Chapter 4

Equations with powers

How do we solve an equation where the unknown is in a power, such as 3ˣ = 20, when trying numbers is slow?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A x = 20 ÷ 3, which is about 6.67.
You picked A. Read on. At the end, the chapter comes back to your guess.
B x is between 2 and 3.
You picked B. Read on. At the end, the chapter comes back to your guess.
C x is between 6 and 7.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve 3ˣ = 20 by taking logs
Solve 4ˣ − 5(2ˣ) + 4 = 0 as a quadratic in 2ˣ

In chapter 2 you solved 2ˣ = 64 by counting powers. For 3ˣ = 20 that fails, because 20 is not a power of 3. We need a tool that brings the power x down to the ground. The power law of logarithms does exactly this: log (aˣ) = x log a.

The method for aˣ = b has three moves. First take logs of both sides. This is allowed because both sides are above 0, and two equal numbers have equal logs. Second use the power law to bring x down. Third divide by log a. The answer is x = log b ÷ log a, with the logs to any one base. A calculator log key uses base 10.

Worked example

Solve 3ˣ = 20. Give x to 3 significant figures.

Estimate first. 3² = 9 and 3³ = 27, and 20 is between them. So x is between 2 and 3.
Take logs of both sides: log (3ˣ) = log 20.
The power law brings x down: x log 3 = log 20.
Divide both sides by log 3: x = log 20 ÷ log 3.
Use log 20 = 1.30103 and log 3 = 0.47712. Then x = 1.30103 ÷ 0.47712 = 2.7268.
Round once, at the end. To 3 significant figures, x = 2.73. This agrees with the estimate.

Answer: x = 2.73

The solution of 3ˣ = 20 is the x value where the curve y = 3ˣ meets the line y = 20.

Often the unknown is in a longer power, or in a power on both sides. The moves are the same. Take logs, bring the powers down, and collect every term in x on one side. Keep the brackets: a power such as 2x − 1 must be multiplied by log 3 as a whole.

Worked example

Solve 3²ˣ⁻¹ = 7ˣ. Give x to 3 significant figures.

Take logs of both sides: log (3²ˣ⁻¹) = log (7ˣ).
Bring the powers down. The whole power 2x − 1 multiplies log 3: (2x − 1) log 3 = x log 7.
Expand the bracket: 2x log 3 − log 3 = x log 7.
Collect the x terms on one side: 2x log 3 − x log 7 = log 3.
Factorise: x (2 log 3 − log 7) = log 3. So x = log 3 ÷ (2 log 3 − log 7).
Use log 3 = 0.47712 and log 7 = 0.84510. Then 2 log 3 − log 7 = 0.95424 − 0.84510 = 0.10914.
Divide: x = 0.47712 ÷ 0.10914 = 4.3716. To 3 significant figures, x = 4.37.

Answer: x = 4.37

Be careful with an inequality. A log to a base above 1 keeps the order, so if one number is bigger, its log is bigger. But the log of a number between 0 and 1 is negative. When you divide both sides by a negative number, the inequality sign flips.

Worked example

A sample has a mass of 1 kg and halves every hour. After x hours its mass is 0.5ˣ kg. Find when the mass is below 0.1 kg. Give the smallest whole number of hours.

We need 0.5ˣ < 0.1.
Take logs of both sides: x log 0.5 < log 0.1.
Now log 0.5 = −0.30103 and log 0.1 = −1. Both are negative.
Divide by log 0.5, which is negative. The sign flips: x > log 0.1 ÷ log 0.5.
To 3 significant figures, x > −1 ÷ −0.30103 = 3.32.
The smallest whole number above 3.32 is 4. Check: 0.5⁴ = 0.0625, which is below 0.1, and 0.5³ = 0.125, which is not.

Answer: x > 3.32, so the smallest whole number of hours is 4.

The mass is below 0.1 kg to the right of the point A, (3.32, 0.1), where the curve is under the line.

Some equations look new but are a quadratic in disguise. In 4ˣ − 5(2ˣ) + 4 = 0 the base 4 is 2². So 4ˣ = (2²)ˣ = (2ˣ)². Both powers are now powers of 2ˣ. Write u for 2ˣ and the equation becomes a quadratic in u. Solve for u, then go back to x. One fact decides which roots you keep: 2ˣ is always above 0, so u must be above 0.

Worked example

Solve 4ˣ − 5(2ˣ) + 4 = 0.

Write 4ˣ as (2ˣ)². Let u = 2ˣ. The equation becomes u² − 5u + 4 = 0.
Factorise: (u − 1)(u − 4) = 0. So u = 1 or u = 4.
Both are above 0, so both can be 2ˣ. Go back to x.
Because 2⁰ = 1, the equation 2ˣ = 1 gives x = 0.
Because 2² = 4, the equation 2ˣ = 4 gives x = 2.
Check x = 0: 4⁰ − 5(2⁰) + 4 = 1 − 5 + 4 = 0. Check x = 2: 16 − 20 + 4 = 0. Both work.

Answer: x = 0 or x = 2.

The curve y = 4ˣ − 5(2ˣ) + 4 meets the x axis at A, (0, 0), and B, (2, 0). These are the two solutions, x = 0 and x = 2.

The power may be shifted, as in 2²ˣ⁺¹. Split it with the law of indices: 2²ˣ⁺¹ = 2²ˣ × 2¹ = 2 × (2ˣ)². The 2 in front changes the quadratic, so it may not factorise as simply. The same rules apply: u = 2ˣ must be above 0.

Worked example

Solve 2²ˣ⁺¹ − 5(2ˣ) + 2 = 0.

Split the power: 2²ˣ⁺¹ = 2²ˣ × 2¹ = 2 × (2ˣ)².
Let u = 2ˣ. The equation becomes 2u² − 5u + 2 = 0.
Factorise: (2u − 1)(u − 2) = 0. So u = 1/2 or u = 2.
Both are above 0, so both can be 2ˣ.
Because 2⁻¹ = 1/2, the equation 2ˣ = 1/2 gives x = −1.
Because 2¹ = 2, the equation 2ˣ = 2 gives x = 1.
Check x = 1: 2³ − 5(2) + 2 = 8 − 10 + 2 = 0. Check x = −1: 2⁻¹ − 5(2⁻¹) + 2 = 0.5 − 2.5 + 2 = 0.

Answer: x = −1 or x = 1.

Worked example

Solve 9ˣ − 4(3ˣ) − 5 = 0. Give any answer that is not exact to 3 significant figures.

Write 9ˣ as (3ˣ)². Let u = 3ˣ. Then u² − 4u − 5 = 0.
Factorise: (u − 5)(u + 1) = 0. So u = 5 or u = −1.
Now u = 3ˣ, and 3ˣ is always above 0. So u = −1 has no solution for x.
Only u = 5 is left: 3ˣ = 5.
Take logs: x log 3 = log 5, so x = log 5 ÷ log 3 = 0.69897 ÷ 0.47712 = 1.46498.
To 3 significant figures, x = 1.46, rounded once from 1.46498. There is only one solution.

Answer: x = 1.46

Equations with logs in them need a different set of checks. Use the laws to write one log, change to index form, and solve. Then test every answer: each number inside an original log must be above 0. An answer that makes a number inside a log zero or negative is an extraneous root, and it must be thrown away.

Worked example

Solve log₂ x + log₂ (x − 2) = 3.

Both logs are to the base 2, so use the product law: log₂ [x(x − 2)] = 3.
Change to index form: x(x − 2) = 2³ = 8.
Expand and collect: x² − 2x − 8 = 0.
Factorise: (x − 4)(x + 2) = 0. So x = 4 or x = −2.
Test the answers in the original equation. We need x above 0 and x − 2 above 0, so x must be above 2.
The value x = −2 breaks both rules, so reject it. For x = 4: log₂ 4 + log₂ 2 = 2 + 1 = 3. So x = 4.

Answer: x = 4

Three more kinds of equation with logs come up in exams. When the bases differ, change them to one base. When a log appears as a square and on its own, call it y. When two logs are tied in a pair of equations, use the product and quotient laws to turn the pair into two ordinary equations. In every case, test each answer against the rule that numbers inside a log must be above 0.

Worked example

Solve log₂ x = log₄ (x + 6).

The bases differ, so change log₄ to base 2: log₄ (x + 6) = log₂ (x + 6) ÷ log₂ 4.
Since 2² = 4, log₂ 4 = 2. So log₄ (x + 6) = 1/2 log₂ (x + 6).
The equation becomes log₂ x = 1/2 log₂ (x + 6). Multiply both sides by 2: 2 log₂ x = log₂ (x + 6).
The power law makes the left side log₂ (x²). So log₂ (x²) = log₂ (x + 6).
Equal logs to the same base mean equal numbers inside: x² = x + 6, so x² − x − 6 = 0.
Factorise: (x − 3)(x + 2) = 0. So x = 3 or x = −2.
We need x above 0 for log₂ x. So reject x = −2. Check x = 3: log₂ 3 = 1.585 and log₄ 9 = 1.585. So x = 3.

Answer: x = 3

Worked example

Solve (log₂ x)² − 3 log₂ x + 2 = 0.

The log appears as a square and on its own. Let y = log₂ x. The equation becomes y² − 3y + 2 = 0.
Factorise: (y − 1)(y − 2) = 0. So y = 1 or y = 2.
Go back to x. If log₂ x = 1, then x = 2¹ = 2.
If log₂ x = 2, then x = 2² = 4.
Both are above 0, so both are allowed. Check x = 4: (log₂ 4)² − 3 log₂ 4 + 2 = 4 − 6 + 2 = 0.

Answer: x = 2 or x = 4.

Worked example

Solve log₃ x + log₃ y = 4 and log₃ x − log₃ y = 2 together.

Use the product law on the first equation: log₃ (xy) = 4. So xy = 3⁴ = 81.
Use the quotient law on the second: log₃ (x ÷ y) = 2. So x ÷ y = 3² = 9, which gives x = 9y.
Put x = 9y into xy = 81: 9y² = 81. So y² = 9.
Then y = 3 or y = −3. A log needs y above 0, so y = 3.
Then x = 9 × 3 = 27. Check: log₃ 27 + log₃ 3 = 3 + 1 = 4, and log₃ 27 − log₃ 3 = 3 − 1 = 2.

Answer: x = 27 and y = 3.

See it three ways

With things

A pile of 1 seed doubles every day: 1, 2, 4, 8, 16. The pile passes 12 seeds between day 3 (8 seeds) and day 4 (16 seeds).

As a picture

In words

The pile reaches 12 seeds after a bit more than 3.5 days. The graph shows where the curve meets the line at 12, and logs give that point exactly.

In symbols

2ˣ = 12, so x = log 12 ÷ log 2 = 1.07918 ÷ 0.30103 = 3.58496, which is 3.58 to 2 decimal places.

Three kinds of equation in this chapter and the first move for each.

Try it

Check a solution on a calculator

You need: a scientific calculator, paper and a pencil.

Work out 3 to the power 2.73 with the power key.
Work out 3 to the power 2.727 and 3 to the power 2.7268.
Write the three results to 2 decimal places.

What you will see: The results get closer to 20 as the power gets closer to the exact value of x. A solution found by logs can always be checked this way. Put it back in the equation and see if you get 20.

If you cannot try it, this is what happens.

A wrong idea many people have

To solve 3ˣ = 20, divide: x = 20 ÷ 3. Or write x = log (20 ÷ 3).

Why it feels right: In 3x = 20 you divide by 3. And the log of a quotient is a familiar law, so a log with 20 ÷ 3 inside looks tidy.

The test that breaks it: If x = 20 ÷ 3 = 6.67, then 3ˣ is about 1 500, not 20. If x = log (20 ÷ 3) = 0.824, then 3ˣ is about 2.5, not 20. Neither works. The correct x = 2.73 gives 20.

What is true: Here x is a power, so division by 3 does not undo it. The log undoes it: x = log 20 ÷ log 3, a quotient of two logs. The quotient law of logs is about a difference, not this. Catch the slip by putting your answer back into 3ˣ. Also keep two more figures than you need until the last line, and round once.

A wrong idea many people have

Both roots of the quadratic in u always give a solution for x.

Why it feels right: Both roots solve the quadratic. Handing back only one feels like losing half the answer.

The test that breaks it: In 9ˣ − 4(3ˣ) − 5 = 0 the roots are u = 5 and u = −1. Try 3ˣ = −1. No power of 3 is negative, so there is no real x, and a log of −1 does not exist.

What is true: A root for u must satisfy u = aˣ, which is above 0. Throw away any root with u ≤ 0. The same rule applies to a log equation: reject any answer that puts zero or a negative number inside a log.

A wrong idea many people have

Solving 4ˣ − 5(2ˣ) + 4 = 0 gives only x = 2, because x = 0 is too plain to count.

Why it feels right: The root u = 4 gives the clear answer 2ˣ = 4, x = 2. The root u = 1 gives x = 0, which does not look like a power.

The test that breaks it: Put x = 0 into the equation: 4⁰ − 5(2⁰) + 4 = 1 − 5 + 4 = 0. It works. So x = 0 is a solution, and giving only x = 2 loses it.

What is true: A quadratic in u has two roots, and each positive root gives an x. The value 2⁰ = 1 means x = 0 is a normal answer. Catch the slip by solving for every root of u before going back to x, then checking each in the original equation.

A wrong idea many people have

Dividing by a negative log does not change the inequality sign.

Why it feels right: For positive numbers the sign never changes, and the log sign is easy to forget.

The test that breaks it: For 0.5ˣ < 0.1, take logs: x log 0.5 < log 0.1, where log 0.5 is negative. Without a flip, x < 3.32, and x = 1 would pass. But 0.5¹ = 0.5, which is not below 0.1. With the flip, x > 3.32, and x = 4 gives 0.0625.

What is true: Dividing both sides of an inequality by a negative number reverses it. A base between 0 and 1 gives a negative log. Catch the slip by testing a value of x on each side of your boundary.

Key words

take logs: apply a log to both sides of an equation, to bring a power down with the power law.
quadratic in disguise: an equation with two powers of the same kind, such as 4ˣ and 2ˣ, that becomes a quadratic after a substitution.
extraneous root: an answer found by correct algebra that fails the original equation, such as a value that puts a negative number inside a log.

Back to your guess

The answer is B: x is between 2 and 3.

Since 3² = 9 and 3³ = 27, and 20 lies between 9 and 27, the power x lies between 2 and 3. The value 20 ÷ 3 is wrong because x is a power, not a number to divide by. To find x exactly we take logarithms, which this chapter shows step by step.

Keep it

Copy this into your exercise book.

aˣ = b: take logs, x = log b ÷ log a. Disguised quadratic: let u = aˣ, keep u > 0. Log equations: one log, index form, reject answers that make a number inside a log 0 or less.

Draw: The curve y = 3ˣ, the dashed line y = 20, and the point where they meet with its guide lines.

M5.2.5 Chapter 5

The number e and natural logs

Is there a best base for exponentials and logarithms, and what is the number e?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Raising 1 + 1/n to the power n gives a number that settles near 2.7 as n grows.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Raising 1 + 1/n to the power n gives a number that grows without limit as n grows.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Raising 1 + 1/n to the power n gives a number that reaches 3 and stays there.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find e as the value of (1 + 1/n)ⁿ for large n
Sketch y = eˣ and y = ln x as mirror images
Solve e²ˣ = 7 using natural logs

A culture grows by 100% in one day. If the growth is added once at the end of the day, its mass is multiplied by 2. Now add the growth in n equal steps. Each step adds 1/n of the mass at that moment. After one step the mass is multiplied by 1 + 1/n, and after n steps it is multiplied by (1 + 1/n)ⁿ.

The value of (1 + 1/n)ⁿ to 5 decimal places. It rises, more and more slowly.

More steps give more growth, but the extra gain shrinks fast. The values do not grow forever. They settle on one number. We call it e, so e is the value that (1 + 1/n)ⁿ gets closer to as n becomes very large. To 5 decimal places e = 2.71828. It has endless digits that never repeat, so it is an irrational number, as √2 is.

The curve (1 + 1/n)ⁿ climbs towards the dashed line y = e, which it never reaches.

Worked example

Work out (1 + 1/50)⁵⁰ to 4 decimal places. How far below e is it?

Here n = 50, so 1 + 1/50 = 1.02.
To 4 decimal places, 1.02⁵⁰ = 2.6916.
Take it from e = 2.7183 (to 4 decimal places): 2.7183 − 2.6916 = 0.0267.
So the value is 2.6916, and it is about 0.027 below e.

Answer: 2.6916, which is 0.027 below e.

Why is e so special? Take the graph of y = aˣ and look at its gradient where it crosses the y axis, at (0, 1). For a = 2 the gradient there is about 0.693, which is less than 1. For a = 3 it is about 1.099, which is more than 1. Somewhere between, there is one base whose gradient at (0, 1) is exactly 1. That base is e.

The gradient of the chord from x = 0 to x = 0.0001, which is very close to the gradient of the curve at (0, 1).

There is more. For y = eˣ, the gradient at every point equals the height of the curve there. At (0, 1) the gradient is 1. At (1, e) the gradient is e. We state this as a fact and prove it in the book on differentiation (M5.4). Every exponential aˣ changes at a rate in proportion to its size. For eˣ the rate equals the size, with no extra factor, which keeps the algebra simplest.

The solid curve is y = eˣ. The dashed line y = x + 1 touches the curve at A, (0, 1), where the gradient is 1. The dashed line y = 2.72x touches it at B, (1, e), where the gradient is e.

The function y = eˣ is the exponential function. Its inverse is the natural logarithm, written ln x. So ln x means the log of x to the base e. Everything from chapters 2 and 3 carries over. Since e⁰ = 1 and e¹ = e, we have ln 1 = 0 and ln e = 1. In general ln (eˣ) = x for every x, and e to the power (ln x) is x for x above 0. The laws of logs hold for ln, because ln is a log.

y = eˣ and y = ln x are mirror images in y = x. The asymptote of eˣ is the x axis, y = 0. The asymptote of ln x is the y axis, x = 0. A is (0, 1) and B is (1, 0).

Mirror images swap the roles of x and y, so the domain of one is the range of the other. For y = eˣ the domain is all real x and the range is y > 0. For y = ln x the domain is x > 0 and the range is all real y. The point (0, 1) on eˣ becomes (1, 0) on ln x, and ln x is negative for x between 0 and 1.

Worked example

Solve e²ˣ = 7. Give x to 3 significant figures.

Take the natural log of both sides: ln (e²ˣ) = ln 7.
Since ln of e to the power k is k for any k, the left side is 2x. So 2x = ln 7.
Use ln 7 = 1.9459 to 4 decimal places. Then x = 1.9459 ÷ 2 = 0.97295.
To 3 significant figures, x = 0.973. The exact answer is x = 1/2 ln 7.

Answer: x = 0.973

Worked example

Solve ln (3x − 1) = 2. Give x to 3 significant figures.

Raise e to the power of both sides. Since e to the power (ln y) is y, the left side is 3x − 1. So 3x − 1 = e².
Use e² = 7.3891 to 4 decimal places. Then 3x = 7.3891 + 1 = 8.3891.
Divide by 3: x = 8.3891 ÷ 3 = 2.7964.
To 3 significant figures, x = 2.80.
Check: 3 × 2.80 − 1 = 7.4, which is above 0, so the log exists.

Answer: x = 2.80

Worked example

Solve e²ˣ − 5eˣ + 6 = 0. Give each answer to 3 significant figures.

Write e²ˣ as (eˣ)². Let u = eˣ. The equation becomes u² − 5u + 6 = 0.
Factorise: (u − 2)(u − 3) = 0. So u = 2 or u = 3. Both are above 0, as eˣ always is.
If eˣ = 2, take ln of both sides: x = ln 2 = 0.693.
If eˣ = 3, take ln of both sides: x = ln 3 = 1.10.

Answer: x = ln 2 = 0.693 or x = ln 3 = 1.10.

Worked example

Sketch y = ln (x − 2). Give its domain, its asymptote and where it crosses the x axis.

The graph of y = ln (x − 2) is the graph of y = ln x moved right by 2 units.
The number inside the log must be above 0, so x − 2 > 0. The domain is x > 2.
The y axis, x = 0, was the asymptote of ln x. It moves right by 2, so the asymptote is x = 2.
Cross the x axis: put y = 0. Then ln (x − 2) = 0, so x − 2 = 1 and x = 3.

Answer: Domain x > 2, asymptote x = 2, crosses the x axis at (3, 0).

The graph of y = ln (x − 2), with its asymptote, the dashed line x = 2. A is the point (3, 0), where it crosses the x axis.

Two more things come up in exams. First, the inverse of an exponential function is found with ln. Second, an equation may hold two natural logs, and the laws of logs join them into one. The rule about positive numbers inside a log applies again.

Worked example

f(x) = e²ˣ + 3. Find the inverse function f⁻¹(x), its domain and its range.

Write y = e²ˣ + 3 and make x the subject. Take 3 from both sides: y − 3 = e²ˣ.
Take ln of both sides: ln (y − 3) = 2x. So x = 1/2 ln (y − 3).
Swap the letters. The inverse is f⁻¹(x) = 1/2 ln (x − 3).
The domain of f⁻¹ is the range of f. Since e²ˣ > 0, f(x) > 3. This matches the rule x − 3 > 0 inside the log. So the domain of f⁻¹ is x > 3.
The range of f⁻¹ is the domain of f, which is all real numbers.

Answer: 1/2 ln (x − 3), with domain x > 3 and range all real numbers.

Worked example

Solve ln x + ln (x + 2) = ln 8.

Use the product law on the left side: ln [x(x + 2)] = ln 8.
Equal logs mean equal numbers inside: x(x + 2) = 8. So x² + 2x − 8 = 0.
Factorise: (x + 4)(x − 2) = 0. So x = −4 or x = 2.
We need x above 0 and x + 2 above 0. The value x = −4 puts −4 inside ln x, so reject it.
Check x = 2: ln 2 + ln 4 = ln 8, because 2 × 4 = 8. So x = 2.

Answer: x = 2

See it three ways

With things

A quantity grows so that after x hours it is eˣ times its starting size. After 1 hour it is 2.718 times as big, after 2 hours 7.389 times, after 3 hours 20.086 times.

As a picture

In words

The natural log undoes the exponential. To find how many hours it took to reach a size, take ln of that size. From 7.389 the log goes back to 2.

In symbols

e² = 7.389, so ln 7.389 = 2. Also e³ = 20.086, so ln 20.086 = 3.

Try it

Find natural logs on a calculator

You need: a scientific calculator, paper and a pencil.

Press the ln key, then 1, then equals. Then do ln 2 and ln 10.
Now try ln 0.5. Write the result.
Try ln (−3). Write what the calculator shows.

What you will see: ln 1 = 0, ln 2 = 0.693, ln 10 = 2.303, and ln 0.5 = −0.693. The log of a number between 0 and 1 is negative. For ln (−3) the calculator shows an error, because no power of e is negative.

If you cannot try it, this is what happens.

A wrong idea many people have

e is exactly 2.718, so e² = 2.718 × 2.718 = 7.3875.

Why it feels right: Calculators and books print 2.718, and it looks like a plain decimal.

The test that breaks it: The true value is e = 2.71828..., with digits that never end. Then e² = 7.38906 to 5 decimal places, not 7.3875. The gap is 0.0015 already, and it grows when you raise e to bigger powers.

What is true: e is an irrational number, like π. Use the e key on a calculator and keep the full value in the working. Round only the final answer.

A wrong idea many people have

ln (−3) = −1.10, because ln 3 = 1.10.

Why it feels right: The minus sign seems to move out, as in −3 = −(3).

The test that breaks it: If ln (−3) = −1.10, then e to the power −1.10 would be −3. But e⁻¹·¹⁰ = 0.333, which is above 0. No power of e is negative, so no number x makes eˣ = −3.

What is true: ln x exists only for x > 0, because eˣ is always above 0. Before you take a natural log, check the number inside is above 0. This is the slip that gives a false root in an equation with logs.

A wrong idea many people have

ln (2x) = 2 ln x.

Why it feels right: The power law brings a number from inside to the front, and the 2 looks like it can do the same.

The test that breaks it: Put x = 3. Then ln (2x) = ln 6 = 1.792, but 2 ln x = 2 × 1.0986 = 2.197. The two are different. It is ln (x²) that equals 2 ln x: ln 9 = 2.197.

What is true: The power law needs a power, as in ln (x²). A product inside the log gives a sum: ln (2x) = ln 2 + ln x. Catch the slip by asking: is the 2 multiplying x, or is it the power of x?

A wrong idea many people have

e² means 2 × e, so e² = 5.44.

Why it feels right: In 2x the number multiplies. A letter with a small 2 beside it looks like the same idea.

The test that breaks it: The small 2 is a power. e² = e × e = 7.389, using the e key. But 2 × e = 5.437. The two numbers differ, and ln 7.389 = 2, while ln 5.437 = 1.693.

What is true: In e², e is the base and 2 is the power. The power says how many e are multiplied together. A multiplier would be written 2e.

Key words

e: the number that (1 + 1/n)ⁿ gets closer to as n grows. e = 2.71828 to 5 decimal places, and it is irrational.
natural logarithm: the log to the base e, written ln x. It is the inverse of eˣ.
limit: the number a list of values gets closer and closer to, as n grows, such as e for (1 + 1/n)ⁿ.

Back to your guess

The answer is A: Raising 1 + 1/n to the power n gives a number that settles near 2.7 as n grows.

The base 1 + 1/n gets closer to 1, which pulls the answer down, while the power n gets bigger, which pushes it up. The two effects balance. The values settle near 2.718. That number is called e, and it is the most useful base in mathematics.

Keep it

Copy this into your exercise book.

e is about 2.71828, the limit of (1 + 1/n)ⁿ. ln x is the log to base e and undoes eˣ. ln (eˣ) = x. The graphs are mirror images in y = x.

Draw: The curves y = eˣ and y = ln x with the line y = x between them, A at (0, 1) and B at (1, 0).

M5.2.6 Chapter 6

Growth and decay models

A herd of goats grows by 8% every year. How do we write a rule for its size, and when does such a rule stop being true?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A About 90 goats, because 4 goats are added each year.
You picked A. Read on. At the end, the chapter comes back to your guess.
B About 540 goats, because 50 × 1.08 × 10 = 540.
You picked B. Read on. At the end, the chapter comes back to your guess.
C About 108 goats, because each year the herd is multiplied by 1.08.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Model a goat herd growing by 8% a year with an exponential
Find a half life from a decay model
Say when an exponential model stops making sense

A herd of 50 goats grows by 8% each year. The size after one year is 50 × 1.08, because 100% + 8% = 108% = 1.08. After two years it is 50 × 1.08 × 1.08. After t years it is 50 multiplied by 1.08, t times. This is a power. The model is N = 50 × 1.08 to the power t, where N is the number of goats and t is the time in years.

The general rule has a starting value A, a rate r for each time unit, and a time t. The size is y = A(1 + r) to the power t. With r = 0.08 the factor 1 + r is 1.08, a growth factor. For a fall of 20% each year, r = −0.2 and the factor is 0.8. A factor below 1 means decay, and a factor above 1 means growth.

Worked example

A herd of 50 goats grows by 8% each year. Find its size after 5 years.

The growth factor each year is 1 + 0.08 = 1.08.
After t years the size is N = 50 × 1.08 to the power t.
Put t = 5. To 4 decimal places, 1.08 to the power 5 = 1.4693.
N = 50 × 1.4693 = 73.47. A herd has a whole number of goats, so the size is about 73.

Answer: About 73 goats.

The herd N = 50 × 1.08 to the power t. At t = 5 it is 73.47 goats, about 73.

The same growth can be written with e. Papers print e with a raised kt. This book cannot raise a letter, so it writes exp(kt) for the same thing. In general exp(x) means eˣ, so exp(kt) is e to the power kt. Every positive number b equals exp(ln b). So 1 + r = exp(k), where k = ln (1 + r). Then (1 + r) to the power t is exp(kt), and the model becomes y = A exp(kt). The number k is the continuous growth rate. Note that k is not r: for 8% growth, k = ln 1.08, which is a little below 0.08.

The continuous rate k = ln (1 + r) is always a little below r. The gap grows as r grows.

Worked example

Write N = 50 × 1.08 to the power t in the form N = A exp(kt).

Write 1.08 as exp(ln 1.08).
Then 1.08 to the power t is exp(t ln 1.08).
Use ln 1.08 = 0.0770 to 3 significant figures. So N = 50 × exp(0.0770t).
So A = 50, the starting size, and k = 0.0770.

Answer: A = 50 and k = 0.0770.

Two times matter. The doubling time T is when the size is twice the start. Then A exp(kT) = 2A, so exp(kT) = 2 and T = ln 2 ÷ k. For the herd, T = 0.6931 ÷ 0.0770 = 9.0 years. The doubling time does not depend on A, so a herd of 5 or 5 000 doubles in the same time. For decay, k is negative. The half-life is when half is left: exp(kT) = 1/2 and T = ln 2 ÷ |k|, where |k| is k without its minus sign.

Worked example

The mass of a sample is M = 80 exp(−0.035t) grams, where t is the time in days. Find (a) the starting mass, (b) the mass after 20 days, (c) the half-life, (d) the time to fall to 10 g.

(a) Put t = 0. Then e⁰ = 1, so M = 80. The starting mass is 80 g.
(b) Put t = 20. Then M = 80 exp(−0.7) = 80 × 0.4966 = 39.7 g to 3 significant figures.
(c) The half-life is T = ln 2 ÷ 0.035 = 0.6931 ÷ 0.035 = 19.8 days.
(d) We need 80 exp(−0.035t) = 10. Divide by 80: exp(−0.035t) = 1/8.
Take ln of both sides: −0.035t = ln (1/8) = −2.0794. So t = 2.0794 ÷ 0.035 = 59.4 days.
Check: 59.4 days is three half-lives, 3 × 19.8 = 59.4, and 80 → 40 → 20 → 10 halves three times.

Answer: (a) 80 g, (b) 39.7 g, (c) 19.8 days, (d) 59.4 days.

The mass halves in every 19.8 days. A is (19.8, 40), B is (39.6, 20) and C is (59.4, 10).

Not every quantity decays to nothing. A hot drink cools towards the temperature of the room, and stops there. Its model has a fixed part and a decaying part: θ = A + B exp(−kt). Here A is the temperature it settles at. The term B exp(−kt) is the part that dies away.

Worked example

A hot drink has temperature θ = 20 + 70 exp(−0.05t) °C, with t in minutes. Find (a) the starting temperature, (b) the temperature it approaches, (c) when it reaches 50 °C.

(a) Put t = 0. Then exp(0) = 1, so θ = 20 + 70 = 90 °C.
(b) As t grows, exp(−0.05t) gets close to 0, so θ gets close to 20. The drink cools towards 20 °C and never goes below it.
(c) Put θ = 50: 20 + 70 exp(−0.05t) = 50. Take 20 from both sides: 70 exp(−0.05t) = 30.
Divide by 70: exp(−0.05t) = 3/7. Take ln of both sides: −0.05t = ln (3/7) = −0.8473.
So t = 0.8473 ÷ 0.05 = 16.9 minutes.

Answer: (a) 90 °C, (b) 20 °C, (c) 16.9 minutes.

The drink starts at 90 degrees and cools towards the dashed line at 20 degrees. It reaches 50 degrees after 16.9 minutes.

Often a model is built from data. Two readings of a quantity are enough to find A and k. Put the first reading at t = 0 to get A. Then use the second reading to get k by taking logs.

Worked example

A population of bacteria is 1 200 at t = 0 and 1 900 when t = 3 hours. It follows N = A exp(kt). Find k, and the doubling time.

At t = 0, N = A, so A = 1 200.
At t = 3, 1 900 = 1 200 exp(3k). So exp(3k) = 1 900 ÷ 1 200 = 1.5833.
Take ln of both sides: 3k = ln 1.5833 = 0.4595.
To 3 significant figures, k = 0.4595 ÷ 3 = 0.153 per hour.
The doubling time is ln 2 ÷ k. Keep k unrounded: T = 3 ln 2 ÷ ln 1.5833 = 2.0794 ÷ 0.4595 = 4.525, which is 4.53 hours to 3 significant figures.

Answer: k = 0.153 per hour, and the population doubles every 4.53 hours.

Now ask when a model stops making sense. A model is a rule that fits the data we have. It holds near those data. Far from them it can give nonsense. Exponential growth is the clearest case, because it never slows. In real life food, space, disease and other limits slow the growth, and the curve flattens.

Worked example

The herd model N = 50 × 1.08 to the power t was fitted to counts over the first 5 years. Find N after 100 years. A pasture feeds at most 400 goats. When does the model first pass 400?

After 100 years, N = 50 × 1.08 to the power 100 = 109 988 goats.
That is far more than any pasture can feed. The model is being used far outside its data, which cover only 5 years.
For 400 goats, solve 50 × 1.08 to the power t = 400. Divide by 50: 1.08 to the power t = 8.
Take ln of both sides: t ln 1.08 = ln 8. So t = 2.0794 ÷ 0.0770 = 27.0 years.
After about 27 years the model says 400 goats. From then on it should not be trusted, because the pasture is full.

Answer: N = 109 988 after 100 years, and N reaches 400 after 27.0 years.

The model passes the pasture limit of 400 goats after about 27 years.

See it three ways

With things

The herd starts at 50 goats. Each year it is multiplied by 1.08. After 1 year it is 54, and after 2 years it is 58.32.

As a picture

In words

Each year the herd is multiplied by the same factor, 1.08. Equal steps of time multiply the size by equal factors. This is the mark of an exponential.

In symbols

N = 50 × 1.08 to the power t. For t = 2 this gives 50 × 1.08 × 1.08 = 58.32.

Try it

A half-life with beans

You need: 64 dried beans, a pen, a cup.

Mark one side of each bean with a dot. Count 64 beans.
Shake them in the cup and throw them on a table. Take away every bean that lands dot upwards. Count what is left.
Throw the beans that are left, take away the dot-up ones, and count again. Do this 5 times.

What you will see: About half of the beans are taken away each throw, so the number left falls like 64, 32, 16, 8, 4, 2. Your numbers will differ a little by chance. One throw is the half-life.

If you cannot try it, this is what happens.

A wrong idea many people have

Growth by 8% a year means N = 50 × 8 to the power t, or N = 50 × 0.08 to the power t.

Why it feels right: The number 8 is in the question, and 0.08 is how a calculator wants 8%.

The test that breaks it: Put t = 1. With 8 the herd is 50 × 8 = 400 after one year, which is far more than 8% growth. With 0.08 the herd is 50 × 0.08 = 4, which has shrunk. The right answer after one year is 54.

What is true: The factor is 1 + 0.08 = 1.08: the whole herd (100%) plus 8% more. Catch the slip by putting t = 1 into your model and checking it gives 8% more than the start.

A wrong idea many people have

For 8% growth, k = 0.08 in N = A exp(kt).

Why it feels right: Both r and k are called a rate, and 0.08 is the number in the question.

The test that breaks it: With k = 0.08, after one year the factor is exp(0.08) = 1.0833, which is 8.33% growth, not 8%. The right value is k = ln 1.08 = 0.0770, and exp(0.0770) = 1.08.

What is true: The rate r goes with the form (1 + r) to the power t, and k = ln (1 + r) goes with the form exp(kt). They are close for small r but not equal. Convert with k = ln (1 + r).

A wrong idea many people have

After two half-lives everything has gone.

Why it feels right: One half-life takes away half, and two halves make a whole.

The test that breaks it: Take 80 g. After one half-life, 40 g is left. After the second half-life, half of 40 g goes, and 20 g is left. After the third, 10 g.

What is true: Each half-life halves what is left, not what you started with. The amount gets smaller and smaller but never reaches 0. After n half-lives the fraction left is (1/2) to the power n.

A wrong idea many people have

If the model fits the first 5 years well, it will fit 100 years.

Why it feels right: A good fit looks like proof that the rule is the real one.

The test that breaks it: The herd model fitted to 5 years gives 109 988 goats after 100 years. No pasture can feed that many, so the model is wrong there.

What is true: A model is trusted near the data it was built from. Far outside them, things the model ignores, such as food, space and disease, matter. Always say how far a model is used, and ask if the rate can really stay the same.

Key words

growth factor: the number you multiply by each time step, 1 + r. A factor above 1 is growth, and below 1 is decay.
doubling time: the time for a growing quantity to become twice as big. T = ln 2 ÷ k.
half-life: the time for a decaying quantity to fall to half. T = ln 2 ÷ |k|.
model: a rule that fits data and predicts new values, such as N = 50 × 1.08 to the power t.

Back to your guess

The answer is C: About 108 goats, because each year the herd is multiplied by 1.08.

The herd starts at 50 and each year it grows by 8% of its size at that time. So the number added grows too. After 10 years the size is 50 × 1.08 ten times, which is about 108. Adding 4 goats a year would give 90, and 540 comes from multiplying by 10 instead of using a power.

Keep it

Copy this into your exercise book.

Growth by r: y = A(1 + r) to the power t. Also y = A exp(kt), with k = ln (1 + r). Doubling time ln 2 ÷ k. Half-life ln 2 ÷ |k|. Trust a model near its data only.

Draw: A decay curve with dots at each half-life, the mass halving every time.

M5.2.7 Chapter 7

Straight lines from logs

Data that rise in a curve are hard to read. How can logs turn them into a straight line, and what do the gradient and the intercept tell us?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A A straight line with gradient 2.
You picked A. Read on. At the end, the chapter comes back to your guess.
B A curve, because y = 3x² is not a straight line.
You picked B. Read on. At the end, the chapter comes back to your guess.
C A straight line with gradient 3.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Turn y = axⁿ into a straight line by taking logs
Find a and n from a log graph of real data

A straight line is the easiest graph to work with. You can draw it with a ruler, and its gradient and intercept hold the constants of the rule. A curve y = axⁿ or y = A bˣ has two constants, but they are hard to read from a curve. Logs bend the curve into a line. In this chapter lg means log to the base 10, which keeps graph labels short.

Start with y = axⁿ. Take lg of both sides. The product law gives lg y = lg a + lg (xⁿ). The power law gives lg (xⁿ) = n lg x. So lg y = n lg x + lg a. A paper may use ln on both axes instead. Then the gradient is still n and the intercept is ln a. Compare this with Y = mX + c. If Y stands for lg y and X for lg x, then the gradient m is n and the intercept c is lg a. A graph of lg y against lg x is a straight line. It is called a log-log graph.

Worked example

y = 5x³. Show that lg y against lg x is a straight line, and give its gradient and intercept.

Take lg of both sides: lg y = lg (5x³).
The product law splits the right side: lg (5x³) = lg 5 + lg (x³).
The power law brings the 3 down: lg (x³) = 3 lg x.
So lg y = 3 lg x + lg 5. This has the form Y = mX + c, with Y = lg y and X = lg x.
The gradient is m = 3. The intercept is c = lg 5 = 0.699.

Answer: The gradient is 3 and the intercept is 0.699.

See it three ways

With things

The rule y = 3x² gives these pairs. When x = 1, 2, 4 and 8, y = 3, 12, 48 and 192. The points (x, y) climb steeply in a curve.

As a picture

In words

Plot lg y against lg x and the curve straightens. The gradient is 2, the power of x. The line meets the lg y axis at lg 3 = 0.477.

In symbols

lg y = 2 lg x + lg 3. The points are (0, 0.477), (0.301, 1.079), (0.602, 1.681) and (0.903, 2.283).

Now take an exponential, y = A bˣ. Taking logs gives lg y = lg A + x lg b. In the form with e, y = A exp(kx), where exp(kx) means e to the power kx, taking natural logs gives ln y = ln A + kx. Each is a straight line when the log of y is plotted against x itself, and not against the log of x. This is a log-linear graph. The gradient is k (or lg b), and the intercept on the vertical axis is ln A (or lg A).

Which plot straightens which rule.

How can you tell which rule fits data? Look at the table. If y is multiplied by the same number each time x goes up by the same amount, the rule is exponential. If y is multiplied by the same number each time x is multiplied by the same number, the rule is a power law. Or just try both plots and see which is straight.

Here are real data. The Sun has planets that travel in ellipses. For each planet, d is its mean distance from the Sun, in AU. One AU is the mean distance from the Earth to the Sun. T is the time for one orbit, in years. Does T depend on d by a power law T = a dⁿ?

Mean distance from the Sun and time of one orbit for six planets.

The logs of the same readings, to 3 decimal places.

lg T against lg d for the six planets in the table, Mercury lowest and Saturn highest. The straight line is the line of best fit. The points lie on it, so T = a dⁿ is a good rule.

Worked example

Use the planet data to find a and n in T = a dⁿ. Give both to 3 significant figures.

If T = a dⁿ, then lg T = n lg d + lg a. So plot lg T against lg d, as in the table and the graph.
The six points lie very close to one straight line. Draw the line of best fit. A line drawn by eye gives a slightly different gradient, for example 1.48 to 1.52. Any value in that range earns the mark.
Take two points far apart on the line: Mercury at (−0.412, −0.618) and Saturn at (0.979, 1.469).
The gradient is (1.469 − (−0.618)) ÷ (0.979 − (−0.412)) = 2.087 ÷ 1.391 = 1.50.
So n = 1.50. The line passes through the origin: Earth has d = 1 and T = 1, and lg 1 = 0.
The intercept is lg a = 0, so a = 1.00.
So T = d to the power 1.5, with T in years and d in AU. Squaring both sides gives T² = d³, which is Kepler’s third law.

Answer: n = 1.50 and a = 1.00.

The same method works for exponentials, with ln y against x. Here are counts from a culture of bacteria. The numbers are chosen for this example. The time t is in hours and N is the number of cells.

Counts of a culture, and the natural logs of the counts.

Worked example

The counts in the table follow N = A exp(kt). Use the line of ln N against t to find k and A to 3 significant figures.

Take ln of both sides: ln N = ln A + kt. Plot ln N against t. The gradient will be k.
The points lie close to a straight line. Draw the line of best fit. A line drawn by eye gives a slightly different gradient, for example 0.34 to 0.36. Any value in that range earns the mark.
The line passes through (0, 5.302) and (5, 7.046).
The gradient is k = (7.046 − 5.302) ÷ 5.
So k = 1.744 ÷ 5 = 0.349 per hour.
The intercept is ln A = 5.302. So A = exp(5.302) = 201, to 3 significant figures.

Answer: k = 0.349 per hour and A = 201.

ln N against t for the culture. The dots are the six readings and the straight line is the line of best fit.

You can also find the constants from just two exact points, without a graph. The two readings give two equations in ln a and n. Subtract them to cancel ln a, and you are left with an equation for n.

Worked example

Two variables are related by y = a xⁿ. When x = 2, y = 12. When x = 8, y = 96. Find n and a.

Take ln of both sides: ln y = ln a + n ln x. Use each reading.
From (2, 12): ln 12 = ln a + n ln 2. From (8, 96): ln 96 = ln a + n ln 8.
Subtract the first from the second: ln 96 − ln 12 = n (ln 8 − ln 2).
The quotient law gives ln 8 = n ln 4. Since 8 = 2³ and 4 = 2², this is 3 ln 2 = 2n ln 2. So n = 1.5.
Put n = 1.5 in the first reading: 12 = a × 2 to the power 1.5 = a × 2.828. So a = 12 ÷ 2.828 = 4.24.

Answer: n = 1.5 and a = 4.24.

Worked example

A graph of lg y against x is a straight line through the points (1, 0.9) and (4, 1.8). The rule is y = A bˣ. Find A and b to 3 significant figures.

Take lg of both sides: lg y = lg A + x lg b. So the gradient is lg b and the intercept on the vertical axis is lg A.
The gradient is (1.8 − 0.9) ÷ (4 − 1) = 0.3. So lg b = 0.3.
Undo the lg: b = 10 to the power 0.3 = 2.00 to 3 significant figures.
The line is given by two points that are not on the vertical axis, so find its height at x = 0. Going back 1 step from x = 1 lowers lg y by 0.3. So lg A = 0.9 − 0.3 = 0.6.
Undo the lg. To 3 significant figures, A = 10 to the power 0.6 = 3.98.

Answer: A = 3.98 and b = 2.00.

Worked example

A graph of ln y against x is a straight line with gradient 0.5 that passes through (0, 1.2). Write y in the form A exp(kx), and find A and k to 3 significant figures.

The line is ln y = 0.5x + 1.2, with gradient 0.5 and intercept 1.2.
Compare with ln y = ln A + kx. So k = 0.5 and ln A = 1.2.
Undo the ln: A = exp(1.2) = 3.32.
So y = 3.32 exp(0.5x).

Answer: A = 3.32 and k = 0.5.

Try it

Straighten a curve yourself

You need: a scientific calculator, paper and a pencil.

The rule is y = 2x². Work out y for x = 1, 2, 3 and 4.
Work out lg x and lg y for each pair. Write them to 3 decimal places.
Find the gradient from the first pair to the last pair: the change in lg y divided by the change in lg x.

What you will see: The gradient is 2, the power of x. The line starts at lg 2 = 0.301 when lg x = 0. The numbers in lg y rise by equal amounts for equal steps in lg x, which makes the straight line.

If you cannot try it, this is what happens.

A wrong idea many people have

On the graph of lg y against lg x, the intercept is a.

Why it feels right: In y = mx + c the intercept is a constant you can read off. In y = axⁿ the constant is a, so it looks like the intercept.

The test that breaks it: For y = 5x³ the line is lg y = 3 lg x + lg 5. The intercept is lg 5 = 0.699, not 5. At the point lg x = 0, the height is 0.699, and 10 to the power 0.699 is 5.

What is true: The intercept is lg a, so a = 10 to the power (intercept). For the natural log version, a = exp(intercept). Catch the slip by asking: is the intercept a log, or the number itself?

A wrong idea many people have

To find n in y = axⁿ, plot lg y against x and take the gradient.

Why it feels right: The exponential rule gives a straight line when you plot lg y against x, and the plot is simpler.

The test that breaks it: For y = 3x², plot lg y against x: the points (1, 0.477), (2, 1.079), (4, 1.681), (8, 2.283) are not a line. The gradient from the first pair is 0.60, and from the last pair 0.15.

What is true: A power law needs both axes in logs. An exponential needs only y in logs. If a plot is not straight, you have the wrong plot for the rule. Try the other one.

A wrong idea many people have

You can take lg of y and ln of x on the same graph.

Why it feels right: Both are logarithms, and a log is a log.

The test that breaks it: For y = x², lg y = 2 lg x. Plot lg y against ln x. Since ln x = 2.3026 lg x, the gradient is 2 ÷ 2.3026 = 0.8686, not 2.

What is true: Use the same base on both axes of a log-log graph. Then the gradient is n. With different bases, the gradient is n times a conversion factor. Choose one base and keep it.

A wrong idea many people have

y = 3x² + 4 gives a straight line when you plot lg y against lg x.

Why it feels right: Taking lg of both sides and using the product law seems to split the sum into lg (3x²) + lg 4.

The test that breaks it: The product law splits a product, not a sum. For x = 1 and x = 2, y = 7 and 16, with lg y = 0.845 and 1.204 and lg x = 0 and 0.301. The gradient is 1.19. For x = 4 and 8, y = 52 and 196. The gradient there is 1.91. The gradient changes, so it is not a line.

What is true: A constant added to axⁿ ruins the log trick, because lg (a + b) is not lg a + lg b. Take the constant away first: plot lg (y − 4) against lg x to get a line with gradient 2.

Key words

log-log graph: a graph of the log of y against the log of x. A power law y = axⁿ gives a straight line.
log-linear graph: a graph of the log of y against x. An exponential law gives a straight line.
line of best fit: a straight line drawn through the middle of a set of points, with about as many points on each side.

Back to your guess

The answer is A: A straight line with gradient 2.

If y = 3x², then lg y = lg 3 + 2 lg x. This is a straight line when lg y is plotted against lg x. Its gradient is the power 2, and its intercept is lg 3. The 3 is not the gradient. It hides in the intercept, as lg 3.

Keep it

Copy this into your exercise book.

For y = axⁿ, plot lg y against lg x. The gradient is n and the intercept is lg a. For y = A exp(kx), plot ln y against x. The gradient is k. Use one log base.

Draw: A straight line on log axes with its gradient triangle and its intercept marked.

The graph of y = 3ˣ

Watch 2ˣ fill the gaps

The graph of y = log₂ x

Find logs on a calculator

Test the product law

Check a solution on a calculator

Find natural logs on a calculator

A half-life with beans

Straighten a curve yourself

Key words, A to Z