Chapter 2 of 7

All 7 chapters
  1. 1Exponential functions, readunderstood
  2. 2Logarithms, readunderstood
  3. 3Laws of logarithms, readunderstood
  4. 4Equations with powers, readunderstood
  5. 5The number e and natural logs, readunderstood
  6. 6Growth and decay models, readunderstood
  7. 7Straight lines from logs, readunderstood

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

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • 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.

Two lines of symbols. The top line reads 2, ⁵, =, 32 with the words base over the 2, power over the ⁵ and number over the 32. The bottom line reads log₂, 32, =, 5 with the words base under log₂, number under 32 and power under 5. Three arrows run down: the base goes to log₂, the power goes to the 5, and the number goes to the 32 inside the log.basepowernumber2⁵=32log₂32=5basenumberpower
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.

Two lines of symbols. The top line reads 2, ⁷, =, 128 with the words base over the 2, power over the ⁷ and number over the 128. The bottom line reads log₂, 128, =, 7 with the words base under log₂, number under 128 and power under 7. Three arrows run down: the base goes to log₂, the power goes to the 7, and the number goes to the 128 inside the log.basepowernumber2⁷=128log₂128=7basenumberpower
The switch for 2⁷ = 128.

Worked example

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

  1. In 2⁷ = 128 the base is 2, the power is 7 and the number is 128.
  2. The log of the number to the base is the power: log₂ 128 = 7.
  3. In log₇ 49 = 2 the base is 7, the number is 49 and the answer, 2, is the power.
  4. 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.

  1. log₃ 81 asks: 3 to which power gives 81?
  2. Count the powers of 3: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81.
  3. So log₃ 81 = 4.
  4. log₂ (1/8) asks: 2 to which power gives 1/8?
  5. A negative power gives a fraction. Since 2³ = 8, we have 2⁻³ = 1/8.
  6. 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.

A table with the heads x and log₂ x. Rows: 1/8 and −3; 1/4 and −2; 1/2 and −1; 1 and 0; 2 and 1; 4 and 2; 8 and 3.xlog₂ x1/8−31/4−21/2−110214283
Values of log₂ x. Each time x doubles, log₂ x goes up by 1.

Draw it first

The graph of y = log₂ x

  1. A graph with the x axis x and the y axis y. Step 1 plots five dots: (1/2, −1), (1, 0), (2, 1), (4, 2) and (8, 3). Step 2 joins them with one smooth curve that climbs steeply near the y axis and then rises more and more slowly.12345678−2−11230xy

    1 Plot the points (x, log₂ x) from the table.

  2. A graph with the x axis x and the y axis y. Step 1 plots five dots: (1/2, −1), (1, 0), (2, 1), (4, 2) and (8, 3). Step 2 joins them with one smooth curve that climbs steeply near the y axis and then rises more and more slowly.12345678−2−11230xy

    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.

A graph with the x axis x and the y axis y, with equal units on both axes. A solid curve named y = 2ˣ rises steeply. A solid curve named y = log₂ x rises slowly from the lower left. A dashed straight line named y = x runs between them at 45 degrees. The two curves are mirror images in the dashed line. The point A is (3, 8) on the first curve and the point B is (8, 3) on the second.ABy = xy = 2ˣy = log₂ x2345678923456789xy
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.

  1. (a) Change to index form: x = 5³ = 125.
  2. (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.
  3. (c) Change to index form: 3x + 1 = 2⁴ = 16.
  4. Take 1 from both sides: 3x = 15. Divide by 3: x = 5.
  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

A table with the heads doublings and seeds. Rows: 0 and 1; 1 and 2; 2 and 4; 3 and 8; 4 and 16; 5 and 32.doublingsseeds01122438416532

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.

  1. Press the log key, then 10, to find log 10.
  2. Find log 100, then log 1000.
  3. 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 table with the heads x and log x. Rows: 10 and 1; 100 and 2; 1 000 and 3; 50 and 1.699; 5 000 and 3.699.xlog x10110021 0003501.6995 0003.699

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.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which of these is 5³ = 125 written as a logarithm?

    Tap your answer.

    1. Not this one. You used the power 3 as the base. In the log form the base stays the base, 5, and the power 3 becomes the answer. Go to M5.2.2. Try another.

    2. Not this one. You put the answer and the number the wrong way round. The number, 125, goes inside the log and the power, 3, is the answer. Go to M5.2.2. Try another.

    3. Not this one. You used the number 125 as the base. The base stays 5, and 125 goes inside the log. Go to M5.2.2. Try another.

    4. Yes, that is right. The base is 5, the power is 3 and the number is 125. In log form the base stays the base, the number goes inside the log, and the power is the answer.

    See the answer

    log₅ 125 = 3

    The base is 5, the power is 3 and the number is 125. In log form the base stays the base, the number goes inside the log, and the power is the answer.

    • If you chose A: You used the power 3 as the base. In the log form the base stays the base, 5, and the power 3 becomes the answer. Go to M5.2.2.
    • If you chose B: You put the answer and the number the wrong way round. The number, 125, goes inside the log and the power, 3, is the answer. Go to M5.2.2.
    • If you chose C: You used the number 125 as the base. The base stays 5, and 125 goes inside the log. Go to M5.2.2.

  2. What is log₄ 64?

    Tap your answer.

    1. Yes, that is right. Count powers of 4: 4¹ = 4, 4² = 16, 4³ = 64. So log₄ 64 = 3.

    2. Not this one. You divided 64 by 4. But a log is a power: it asks 4 to which power gives 64. Go to M5.2.2. Try another.

    3. Not this one. You multiplied 64 by 4. A log is the power, not a product. Go to M5.2.2. Try another.

    4. Not this one. You may have taken the square root of 64. The question is the power of 4 that gives 64: 4¹ = 4, 4² = 16, 4³ = 64. Go to M5.2.2. Try another.

    See the answer

    3

    Count powers of 4: 4¹ = 4, 4² = 16, 4³ = 64. So log₄ 64 = 3.

    • If you chose B: You divided 64 by 4. But a log is a power: it asks 4 to which power gives 64. Go to M5.2.2.
    • If you chose C: You multiplied 64 by 4. A log is the power, not a product. Go to M5.2.2.
    • If you chose D: You may have taken the square root of 64. The question is the power of 4 that gives 64: 4¹ = 4, 4² = 16, 4³ = 64. Go to M5.2.2.

  3. Which of these does not exist?

    Tap your answer.

    1. Not this one. The number 1/8 is above 0, so it does exist. 2⁻³ = 1/8, so log₂ (1/8) = −3. Go to M5.2.2. Try another.

    2. Not this one. The number 1 is above 0, so this exists. It is 0, since 10⁰ = 1. Go to M5.2.2. Try another.

    3. Yes, that is right. The number inside a log must be above 0, because no power of 5 is negative. The number here is −25, so the logarithm does not exist.

    4. Not this one. This exists. It is 1, since 3¹ = 3. Go to M5.2.2. Try another.

    See the answer

    log₅ (−25)

    The number inside a log must be above 0, because no power of 5 is negative. The number here is −25, so the logarithm does not exist.

    • If you chose A: The number 1/8 is above 0, so it does exist. 2⁻³ = 1/8, so log₂ (1/8) = −3. Go to M5.2.2.
    • If you chose B: The number 1 is above 0, so this exists. It is 0, since 10⁰ = 1. Go to M5.2.2.
    • If you chose D: This exists. It is 1, since 3¹ = 3. Go to M5.2.2.

  4. The log of 9 to the base a is 2. What is a?

    Tap your answer.

    1. Yes, that is right. In index form, a² = 9, so a = 3 or −3. A base must be above 0, so a = 3.

    2. Not this one. It is true that (−3)² = 9, but a base must be above 0. So only a = 3 is allowed. Go to M5.2.2. Try another.

    3. Not this one. You divided 9 by 2. The log form says a² = 9, so a is a square root of 9. Go to M5.2.2. Try another.

    4. Not this one. You squared 9. The index form is a² = 9, so a is the square root of 9, not 9 squared. Go to M5.2.2. Try another.

    See the answer

    3

    In index form, a² = 9, so a = 3 or −3. A base must be above 0, so a = 3.

    • If you chose B: It is true that (−3)² = 9, but a base must be above 0. So only a = 3 is allowed. Go to M5.2.2.
    • If you chose C: You divided 9 by 2. The log form says a² = 9, so a is a square root of 9. Go to M5.2.2.
    • If you chose D: You squared 9. The index form is a² = 9, so a is the square root of 9, not 9 squared. Go to M5.2.2.

  5. Work out log₆ 1 + log₆ 6.

    Tap your answer.

    1. Not this one. You may have thought every log of 1 or of the base is 0. Only log₆ 1 = 0. The other term, log₆ 6, is 1, because 6¹ = 6. Go to M5.2.2. Try another.

    2. Not this one. You added the numbers inside the logs. A log is a power: log₆ 1 is 0 and log₆ 6 is 1. Go to M5.2.2. Try another.

    3. Not this one. You may have taken log₆ 6 as 6. It is 1, because 6¹ = 6, so the sum is 0 + 1 = 1. Go to M5.2.2. Try another.

    4. Yes, that is right. log₆ 1 = 0 because 6⁰ = 1, and log₆ 6 = 1 because 6¹ = 6. The sum is 0 + 1 = 1.

    See the answer

    1

    log₆ 1 = 0 because 6⁰ = 1, and log₆ 6 = 1 because 6¹ = 6. The sum is 0 + 1 = 1.

    • If you chose A: You may have thought every log of 1 or of the base is 0. Only log₆ 1 = 0. The other term, log₆ 6, is 1, because 6¹ = 6. Go to M5.2.2.
    • If you chose B: You added the numbers inside the logs. A log is a power: log₆ 1 is 0 and log₆ 6 is 1. Go to M5.2.2.
    • If you chose C: You may have taken log₆ 6 as 6. It is 1, because 6¹ = 6, so the sum is 0 + 1 = 1. Go to M5.2.2.

  6. For which values of x does log₂ x exist?

    Tap your answer.

    1. Not this one. The value x = 0 is not allowed. No power of 2 gives 0, so log₂ 0 does not exist. Go to M5.2.2. Try another.

    2. Yes, that is right. The log is the inverse of 2ˣ, whose values are all above 0. So log₂ x exists for exactly the x above 0.

    3. Not this one. A power of 2 is never 0 or negative, so log₂ x has no value for x ≤ 0. Go to M5.2.2. Try another.

    4. Not this one. Numbers between 0 and 1 are allowed. For example log₂ (1/2) = −1, because 2⁻¹ = 1/2. Go to M5.2.2. Try another.

    See the answer

    x > 0

    The log is the inverse of 2ˣ, whose values are all above 0. So log₂ x exists for exactly the x above 0.

    • If you chose A: The value x = 0 is not allowed. No power of 2 gives 0, so log₂ 0 does not exist. Go to M5.2.2.
    • If you chose C: A power of 2 is never 0 or negative, so log₂ x has no value for x ≤ 0. Go to M5.2.2.
    • If you chose D: Numbers between 0 and 1 are allowed. For example log₂ (1/2) = −1, because 2⁻¹ = 1/2. Go to M5.2.2.

  7. The graph shows y = log x to some base a. What is a?

    A graph with the x axis x and the y axis y. A curve rises from the lower left, passes through the point (1, 0) and then through the point (9, 2), rising more and more slowly.(9, 2)36912−2−11230xy
    Tap your answer.

    1. Not this one. With base 2 the point with y = 2 would have x = 4, since log₂ 4 = 2. This graph has x = 9. Go to M5.2.2. Try another.

    2. Not this one. The 9 is the number inside the log, not the base. The point (9, 2) says the log of 9 to the base a is 2, so a² = 9. Go to M5.2.2. Try another.

    3. Yes, that is right. The point (9, 2) says the log of 9 to the base a is 2, which means a² = 9. A base is above 0, so a = 3.

    4. Not this one. You divided 9 by 2. The point (9, 2) says a² = 9, so a is a square root of 9. Go to M5.2.2. Try another.

    See the answer

    3

    The point (9, 2) says the log of 9 to the base a is 2, which means a² = 9. A base is above 0, so a = 3.

    • If you chose A: With base 2 the point with y = 2 would have x = 4, since log₂ 4 = 2. This graph has x = 9. Go to M5.2.2.
    • If you chose B: The 9 is the number inside the log, not the base. The point (9, 2) says the log of 9 to the base a is 2, so a² = 9. Go to M5.2.2.
    • If you chose D: You divided 9 by 2. The point (9, 2) says a² = 9, so a is a square root of 9. Go to M5.2.2.

  8. From an earlier chapter

    What is the range of y = 3ˣ − 1?

    Tap your answer.

    1. Not this one. That is the range of y = 3ˣ. Subtracting 1 moves every value down by 1. Go to M5.2.1. Try another.

    2. Yes, that is right. Since 3ˣ is above 0, 3ˣ − 1 is above −1. The curve gets as close to −1 as you like and never reaches it.

    3. Not this one. The curve is above its asymptote y = −1, not below. Since 3ˣ is above 0, 3ˣ − 1 is above −1. Go to M5.2.1. Try another.

    4. Not this one. The asymptote has moved down by 1 to y = −1, not up to y = 1. Go to M5.2.1. Try another.

    See the answer

    y > −1

    Since 3ˣ is above 0, 3ˣ − 1 is above −1. The curve gets as close to −1 as you like and never reaches it.

    • If you chose A: That is the range of y = 3ˣ. Subtracting 1 moves every value down by 1. Go to M5.2.1.
    • If you chose C: The curve is above its asymptote y = −1, not below. Since 3ˣ is above 0, 3ˣ − 1 is above −1. Go to M5.2.1.
    • If you chose D: The asymptote has moved down by 1 to y = −1, not up to y = 1. Go to M5.2.1.

  9. Work out log₂ 64.

    See the answer

    6

    Count the powers of 2: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64. So log₂ 64 = 6.

  10. Solve log₂ (x + 3) = 5.

    See the answer

    x = 29

    In index form, x + 3 = 2⁵ = 32. So x = 32 − 3 = 29. Check: 29 + 3 = 32 is above 0 and log₂ 32 = 5.

  11. Find the base x for which the log of 1/16 to the base x is −2.

    See the answer

    4

    In index form, x to the power −2 is 1/16. So x² = 16, which gives x = 4 or x = −4. A base must be above 0, so x = 4.

  12. Find log₄ 8. Give your answer as a decimal.

    See the answer

    1.5

    Write 4ˣ = 8 with base 2: 4 = 2², so 2²ˣ = 2³. Then 2x = 3 and x = 1.5. Check: 4 to the power 1.5 is 2³ = 8.

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.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (909 KB). It opens on any phone.