Chapter 5 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

M5.2.5 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

  • 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

  • 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.

A table with the heads n and (1 + 1/n)ⁿ. Rows: 1 and 2; 2 and 2.25; 5 and 2.48832; 10 and 2.59374; 100 and 2.70481; 1 000 and 2.71692; 10 000 and 2.71815; 100 000 and 2.71827; 1 000 000 and 2.71828.n(1 + 1/n)ⁿ1222.2552.48832102.593741002.704811 0002.7169210 0002.71815100 0002.718271 000 0002.71828
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.

A graph with the x axis n and the y axis (1 + 1/n)ⁿ. A curve rises from the point (1, 2), steeply at first and then more and more slowly. A dashed horizontal line named y = e lies just above the curve at the right.y = e02040608010022.53n(1 + 1/n)ⁿ
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?

  1. Here n = 50, so 1 + 1/50 = 1.02.
  2. To 4 decimal places, 1.02⁵⁰ = 2.6916.
  3. Take it from e = 2.7183 (to 4 decimal places): 2.7183 − 2.6916 = 0.0267.
  4. 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.

A table with the heads base and gradient at (0, 1). Rows: 2 and 0.693; e and 1.000; 3 and 1.099.basegradient at (0, 1)20.693e1.00031.099
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.

A graph with the x axis x and the y axis y. A solid curve, y = eˣ, rises from the left, through the point A, (0, 1), and the point B, (1, 2.72). A dashed straight line, y = x + 1, touches the curve at A with gradient 1. A second dashed straight line, y = 2.72x, touches the curve at B with gradient 2.72 and passes through the origin.AB−112324680xy
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.

A graph with the x axis x and the y axis y, with equal units on both axes. A solid curve named y = eˣ rises steeply from the left and passes through the point A, (0, 1). A solid curve named y = ln x rises slowly from the lower left and passes through the point B, (1, 0). A dashed line named y = x runs at 45 degrees between them. The curves are mirror images in the dashed line. The x axis is the asymptote of the first curve and the y axis is the asymptote of the second.ABy = xy = eˣy = ln x−112345−1123450xy
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.

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

  1. 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.
  2. Use e² = 7.3891 to 4 decimal places. Then 3x = 7.3891 + 1 = 8.3891.
  3. Divide by 3: x = 8.3891 ÷ 3 = 2.7964.
  4. To 3 significant figures, x = 2.80.
  5. 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.

  1. Write e²ˣ as (eˣ)². Let u = eˣ. The equation becomes u² − 5u + 6 = 0.
  2. Factorise: (u − 2)(u − 3) = 0. So u = 2 or u = 3. Both are above 0, as eˣ always is.
  3. If eˣ = 2, take ln of both sides: x = ln 2 = 0.693.
  4. 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.

  1. The graph of y = ln (x − 2) is the graph of y = ln x moved right by 2 units.
  2. The number inside the log must be above 0, so x − 2 > 0. The domain is x > 2.
  3. The y axis, x = 0, was the asymptote of ln x. It moves right by 2, so the asymptote is x = 2.
  4. 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).

A graph with the x axis x and the y axis y. A curve rises from the bottom, very close to a dashed upright line named x = 2, crosses the x axis at the point A, (3, 0), and rises slowly to the right.Ax = 21345678−4−3−2−11230xy
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.

  1. Write y = e²ˣ + 3 and make x the subject. Take 3 from both sides: y − 3 = e²ˣ.
  2. Take ln of both sides: ln (y − 3) = 2x. So x = 1/2 ln (y − 3).
  3. Swap the letters. The inverse is f⁻¹(x) = 1/2 ln (x − 3).
  4. 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.
  5. 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.

  1. Use the product law on the left side: ln [x(x + 2)] = ln 8.
  2. Equal logs mean equal numbers inside: x(x + 2) = 8. So x² + 2x − 8 = 0.
  3. Factorise: (x + 4)(x − 2) = 0. So x = −4 or x = 2.
  4. We need x above 0 and x + 2 above 0. The value x = −4 puts −4 inside ln x, so reject it.
  5. 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

A table with the heads x and eˣ. Rows: 0 and 1; 1 and 2.718; 2 and 7.389; 3 and 20.086.xeˣ0112.71827.389320.086

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.

  1. Press the ln key, then 1, then equals. Then do ln 2 and ln 10.
  2. Now try ln 0.5. Write the result.
  3. 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 table with the heads x and ln x. Rows: 1 and 0; 2 and 0.693; 10 and 2.303; 0.5 and −0.693; −3 and error.xln x1020.693102.3030.5−0.693−3error

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.

Check yourself

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

  1. Work out ln (e⁵).

    Tap your answer.

    1. Not this one. That is the value of e⁵ itself. The natural log undoes e to a power, so ln (e⁵) is the power, 5. Go to M5.2.5. Try another.

    2. Not this one. That is 5 × e. In e⁵ the 5 is a power, not a multiplier. And ln undoes the power, so it gives back the power 5. Go to M5.2.5. Try another.

    3. Not this one. You may have taken 1 ÷ 5. The natural log of e to a power is that power, which is positive here. Go to M5.2.5. Try another.

    4. Yes, that is right. ln is the inverse of eˣ, so ln (eˣ) = x for every x. With x = 5, ln (e⁵) = 5.

    See the answer

    5

    ln is the inverse of eˣ, so ln (eˣ) = x for every x. With x = 5, ln (e⁵) = 5.

    • If you chose A: That is the value of e⁵ itself. The natural log undoes e to a power, so ln (e⁵) is the power, 5. Go to M5.2.5.
    • If you chose B: That is 5 × e. In e⁵ the 5 is a power, not a multiplier. And ln undoes the power, so it gives back the power 5. Go to M5.2.5.
    • If you chose C: You may have taken 1 ÷ 5. The natural log of e to a power is that power, which is positive here. Go to M5.2.5.

  2. Which line is the asymptote of the graph of y = ln x?

    Tap your answer.

    1. Not this one. The x axis is where the graph crosses, at (1, 0). It is the asymptote of y = eˣ, not of y = ln x. Go to M5.2.5. Try another.

    2. Yes, that is right. The graph of y = ln x is the mirror image of y = eˣ in y = x. The x axis is the asymptote of eˣ, so the y axis, the line x = 0, is the asymptote of ln x.

    3. Not this one. The graph passes through (1, 0) and crosses the line x = 1. The curve never crosses its asymptote. Go to M5.2.5. Try another.

    4. Not this one. The graph crosses y = 1 at x = e. The curve never crosses its asymptote, and ln x keeps rising. Go to M5.2.5. Try another.

    See the answer

    x = 0

    The graph of y = ln x is the mirror image of y = eˣ in y = x. The x axis is the asymptote of eˣ, so the y axis, the line x = 0, is the asymptote of ln x.

    • If you chose A: The x axis is where the graph crosses, at (1, 0). It is the asymptote of y = eˣ, not of y = ln x. Go to M5.2.5.
    • If you chose C: The graph passes through (1, 0) and crosses the line x = 1. The curve never crosses its asymptote. Go to M5.2.5.
    • If you chose D: The graph crosses y = 1 at x = e. The curve never crosses its asymptote, and ln x keeps rising. Go to M5.2.5.

  3. For which values of x does y = ln (x + 3) exist?

    Tap your answer.

    1. Yes, that is right. The number inside a log must be above 0. So x + 3 > 0, which gives x > −3. The graph is y = ln x moved left by 3 units.

    2. Not this one. That is the domain of ln x. Here the number inside is x + 3, so the graph is moved 3 units left. Go to M5.2.5. Try another.

    3. Not this one. The values below −3 make x + 3 negative, which the log cannot take. Go to M5.2.5. Try another.

    4. Not this one. You may have used x − 3 > 0. The number inside the log is x + 3, and it must be above 0. Go to M5.2.5. Try another.

    See the answer

    x > −3

    The number inside a log must be above 0. So x + 3 > 0, which gives x > −3. The graph is y = ln x moved left by 3 units.

    • If you chose B: That is the domain of ln x. Here the number inside is x + 3, so the graph is moved 3 units left. Go to M5.2.5.
    • If you chose C: The values below −3 make x + 3 negative, which the log cannot take. Go to M5.2.5.
    • If you chose D: You may have used x − 3 > 0. The number inside the log is x + 3, and it must be above 0. Go to M5.2.5.

  4. For which values of x is ln (x²) = 2 ln x always true?

    Tap your answer.

    1. Not this one. For x = −3, ln (x²) = ln 9 exists, but 2 ln x needs ln (−3), which does not exist. Go to M5.2.5. Try another.

    2. Not this one. At x = 0, ln x does not exist, so 2 ln x does not exist either. The value 0 must be left out. Go to M5.2.5. Try another.

    3. Yes, that is right. The power law needs the number inside the log to be above 0, and 2 ln x only exists for x > 0. For x below 0, ln (x²) = 2 ln (−x) is the true statement.

    4. Not this one. For x below 0, ln x does not exist, so the right side has no value. Go to M5.2.5. Try another.

    See the answer

    x > 0

    The power law needs the number inside the log to be above 0, and 2 ln x only exists for x > 0. For x below 0, ln (x²) = 2 ln (−x) is the true statement.

    • If you chose A: For x = −3, ln (x²) = ln 9 exists, but 2 ln x needs ln (−3), which does not exist. Go to M5.2.5.
    • If you chose B: At x = 0, ln x does not exist, so 2 ln x does not exist either. The value 0 must be left out. Go to M5.2.5.
    • If you chose D: For x below 0, ln x does not exist, so the right side has no value. Go to M5.2.5.

  5. Mei works out (1 + 1/n)ⁿ for n = 10, 100, 1 000 and 10 000 and gets 2.594, 2.705, 2.717 and 2.718. What does this suggest?

    Tap your answer.

    1. Not this one. The gains get smaller: 0.111, 0.012, 0.001. A sum of shrinking gains can stay below a fixed number. Go to M5.2.5. Try another.

    2. Not this one. The values are already close to 2.72 and rise by tiny amounts. They never get to 3. Go to M5.2.5. Try another.

    3. Not this one. The values are rising, not falling. Each is bigger than the one before. Go to M5.2.5. Try another.

    4. Yes, that is right. The numbers rise, but by smaller and smaller amounts. They are heading for one number, e, which is about 2.718.

    See the answer

    The values rise more and more slowly and settle on a fixed number.

    The numbers rise, but by smaller and smaller amounts. They are heading for one number, e, which is about 2.718.

    • If you chose A: The gains get smaller: 0.111, 0.012, 0.001. A sum of shrinking gains can stay below a fixed number. Go to M5.2.5.
    • If you chose B: The values are already close to 2.72 and rise by tiny amounts. They never get to 3. Go to M5.2.5.
    • If you chose C: The values are rising, not falling. Each is bigger than the one before. Go to M5.2.5.

  6. From an earlier chapter

    Solve 5ˣ = 30. Which is x, to 3 significant figures?

    Tap your answer.

    1. Not this one. You divided 30 by 5. But x is a power: 5⁶ is 15 625, far above 30. Go to M5.2.4. Try another.

    2. Yes, that is right. Take logs: x log 5 = log 30. So x = log 30 ÷ log 5 = 1.47712 ÷ 0.69897 = 2.113, which is 2.11 to 3 significant figures.

    3. Not this one. You worked out log (30 ÷ 5). Taking logs gives x log 5 = log 30, so x = log 30 ÷ log 5, a quotient of two logs. Go to M5.2.4. Try another.

    4. Not this one. You turned the fraction upside down: log 5 ÷ log 30. Since 5² = 25 is below 30, x must be above 2. Go to M5.2.4. Try another.

    See the answer

    2.11

    Take logs: x log 5 = log 30. So x = log 30 ÷ log 5 = 1.47712 ÷ 0.69897 = 2.113, which is 2.11 to 3 significant figures.

    • If you chose A: You divided 30 by 5. But x is a power: 5⁶ is 15 625, far above 30. Go to M5.2.4.
    • If you chose C: You worked out log (30 ÷ 5). Taking logs gives x log 5 = log 30, so x = log 30 ÷ log 5, a quotient of two logs. Go to M5.2.4.
    • If you chose D: You turned the fraction upside down: log 5 ÷ log 30. Since 5² = 25 is below 30, x must be above 2. Go to M5.2.4.

  7. Solve e⁴ˣ = 30. Give x to 3 significant figures.

    See the answer

    0.850

    Take ln of both sides: 4x = ln 30 = 3.4012. So x = 3.4012 ÷ 4 = 0.8503, which is 0.850 to 3 significant figures.

  8. Solve 3e⁻ˣ = 5. Give x to 3 significant figures.

    See the answer

    −0.511

    Divide by 3: e⁻ˣ = 5/3. Take ln of both sides: −x = ln (5/3) = 0.5108. So x = −0.5108, which is −0.511 to 3 significant figures. Check: x is negative, so e⁻ˣ is above 1, and 3e⁻ˣ = 5 is above 3.

  9. Solve e²ˣ − 7eˣ + 10 = 0. Give both answers to 3 significant figures, the smaller first.

    See the answer

    x = 0.693 or x = 1.61

    Let u = eˣ. Then u² − 7u + 10 = 0, which is (u − 2)(u − 5) = 0. So eˣ = 2 gives x = ln 2 = 0.693, and eˣ = 5 gives x = ln 5 = 1.61. Both u are above 0.

  10. Solve ln (2x + 5) = 3. Give x to 3 significant figures.

    See the answer

    7.54

    Raise e to the power of both sides: 2x + 5 = e³ = 20.0855. So 2x = 15.0855 and x = 7.5428, which is 7.54 to 3 significant figures. Check: 2x + 5 = 20.09, which is above 0.

  11. g(x) = e³ˣ − 2. Which of these is the inverse function g⁻¹(x)?

    Tap your answer.

    1. Not this one. You took 2 from both sides, but the 2 in g is taken away. To undo it you add 2: y + 2 = e³ˣ. Go to M5.2.5. Try another.

    2. Not this one. The 3 multiplies x in the power, so it must be divided out after the ln, not multiplied. Go to M5.2.5. Try another.

    3. Not this one. The 3 multiplies x. It is not added to the power, so you divide by 3 after taking the ln. Go to M5.2.5. Try another.

    4. Yes, that is right. Write y = e³ˣ − 2 and add 2: y + 2 = e³ˣ. Take ln: ln (y + 2) = 3x, so x = 1/3 ln (y + 2). Swapping the letters gives g⁻¹(x) = 1/3 ln (x + 2).

    See the answer

    1/3 ln (x + 2)

    Write y = e³ˣ − 2 and add 2: y + 2 = e³ˣ. Take ln: ln (y + 2) = 3x, so x = 1/3 ln (y + 2). Swapping the letters gives g⁻¹(x) = 1/3 ln (x + 2).

    • If you chose A: You took 2 from both sides, but the 2 in g is taken away. To undo it you add 2: y + 2 = e³ˣ. Go to M5.2.5.
    • If you chose B: The 3 multiplies x in the power, so it must be divided out after the ln, not multiplied. Go to M5.2.5.
    • If you chose C: The 3 multiplies x. It is not added to the power, so you divide by 3 after taking the ln. Go to M5.2.5.

  12. Solve ln x + ln (x − 3) = ln 10.

    Tap your answer.

    1. Not this one. The root x = −2 puts −2 inside ln x, and the log of a negative number does not exist. It must be rejected. Go to M5.2.5. Try another.

    2. Not this one. You may have added 3 to 10. The product law gives x(x − 3) = 10, which is a quadratic. Go to M5.2.5. Try another.

    3. Yes, that is right. The product law gives ln [x(x − 3)] = ln 10, so x² − 3x − 10 = 0, which is (x − 5)(x + 2) = 0. We need x above 3, so x = 5. Check: ln 5 + ln 2 = ln 10.

    4. Not this one. This is the root that fails. It makes x negative, and ln x does not exist. Go to M5.2.5. Try another.

    See the answer

    x = 5

    The product law gives ln [x(x − 3)] = ln 10, so x² − 3x − 10 = 0, which is (x − 5)(x + 2) = 0. We need x above 3, so x = 5. Check: ln 5 + ln 2 = ln 10.

    • If you chose A: The root x = −2 puts −2 inside ln x, and the log of a negative number does not exist. It must be rejected. Go to M5.2.5.
    • If you chose B: You may have added 3 to 10. The product law gives x(x − 3) = 10, which is a quadratic. Go to M5.2.5.
    • If you chose D: This is the root that fails. It makes x negative, and ln x does not exist. Go to M5.2.5.

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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).

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

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