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Cambridge IGCSE0606

Logarithmic and exponential functions

Mathematics Additional 0606 Chapter Notes

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Logarithmic and exponential functions
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In this chapter

  1. 1Exponentials and logarithms as inverses
  2. 2Graphs and asymptotes
  3. 3Laws of logarithms
  4. 4Change of base
  5. 5Solving equations with an unknown power
  6. 6Equations that become quadratics
  7. 7Solving logarithmic equations

1. Exponentials and logarithms as inverses

a^(x) = b and log_a b = x say the same thing. A logarithm may be given to any base. ln x means log_e x and lg x means log₁₀ x.

f(x) = e^(x) and g(x) = ln x are each the inverse of the other. Their graphs are a reflection in the line y = x.

So the domain of the inverse is the range of the original, and the range of the inverse is the domain of the original.

a^(x) = b iff x = log_a b

ln(e^(x)) = x e^(ln x) = x

Examiner insight

“Most candidates demonstrated good understanding of the definition of a logarithm and were able to change the exponential form of the equation into the correct logarithmic form.” “Some responses showed that candidates were familiar with the idea that the domain of the inverse related to the range of the original function.”

Worked example 14 marks

A function f is defined by f(x) = e^(2x+1) for x ∈ R.

(a) State the range of f⁻¹.

(b) Find the domain of f⁻¹.

(c) Find an expression for f⁻¹(x).

  1. 1

    (a) The range of f⁻¹ is the domain of f: f⁻¹(x) ∈ R [1]

  2. 2

    (b) The domain of f⁻¹ is the range of f: x > 0 [1]

  3. 3

    (c) y = e^(2x+1), so ln y = 2x + 1 [1]

  4. 4

    x = (ln y - 1)/2, so f⁻¹(x) = (ln x - 1)/2 [1]

2. Graphs and asymptotes

e^(nx) is always positive, so a curve y = ke^(nx) + a gets closer and closer to the asymptote y = a without reaching it.

ln(ax +b) exists only when ax + b > 0. So y = kln(ax +b) has a vertical asymptote where ax + b = 0, at x = -b/a.

Put x = 0 to find the y-intercept and y = 0 to find the x-intercept. Give exact coordinates, and write each asymptote as an equation such as y = -4, not just the number.

y = ke^(nx) + a: asymptote y = a

y = kln(ax + b): asymptote x = -b/a

Examiner insight

“Candidates should be familiar with the asymptotic behaviour of exponential functions.” “Some omitted the asymptotes or did not make them clear.”

Worked example 14 marks

A curve has equation y = 3e^(2x) - 4. State the equation of its asymptote. Find the exact coordinates of the points where the curve crosses the axes.

  1. 1

    e^(2x) > 0, so the curve approaches y = -4: the asymptote is y = -4 [1]

  2. 2

    x = 0: y = 3 - 4 = -1, so the point is (0, -1) [1]

  3. 3

    y = 0: 3e^(2x) = 4, so e^(2x) = 4/3 [1]

  4. 4

    2x = ln4/3, so the point is (1/2ln4/3, 0) [1]

3. Laws of logarithms

For logarithms to the same base, adding logs multiplies and subtracting logs divides. A power comes down in front: the power rule. Also 1 = log_a a.

To write an expression as a single logarithm, turn each number into a log of the same base (3 = lg 1000), then use the laws.

There is no law for the log of a sum or a difference: lg(x -y) is not lg x - lg y.

log_a x + log_a y = log_a (xy)

log_a x - log_a y = log_a x/y

log_a (xⁿ) = nlog_a x

Examiner insight

“Candidates who were unable to deal with the term of 3 and write it as lg1000 were able to gain credit for appropriate use of the power rule and division rule for logarithms.” “A few candidates used logarithms in an invalid way.”

Worked example 13 marks

Write 3 + 2lg p - lg q as a single logarithm to base 10.

  1. 1

    3 = lg 1000 and 2lg p = lg p² [1]

  2. 2

    lg 1000 + lg p² - lg q [1]

  3. 3

    = lg 1000p²/q [1]

4. Change of base

To change a logarithm to a new base c, divide the log of the number by the log of the old base, both to base c. Choose the base so that every log in the question has the same base.

A special case turns a reciprocal into a natural logarithm: 1/(log₅e) = ln 5.

log_a b = (log_c b)/(log_c a)

log_a b = 1/(log_b a)

Examiner insight

“Most candidates changed the logarithms to a consistent base of 2 and were mostly successful.” “A few candidates chose to use either base 10 logarithms or natural logarithms. These candidates were usually less successful as extra calculations involving these logarithms were needed.”

Worked example 12 marks

Write 1/(log_xe) - ln(x - 1) as a single logarithm to base e.

  1. 1

    1/(log_xe) = log_e x = ln x [1]

  2. 2

    ln x - ln(x - 1) = lnx/(x - 1) [1]

5. Solving equations with an unknown power

Take logarithms of both sides, to any base, and use the power rule to bring x down. Divide by any coefficient first.

If the question asks for a decimal, round only at the end. Otherwise leave the answer as an exact logarithm.

a^(x) = b ⇒ xlg a = lg b ⇒ x = (lg b)/(lg a)

Examiner insight

“Most candidates divided by 5 before taking logarithms of both sides. This resulted in the easiest method of solution.” “Candidates did not always evaluate their final answer and there were some who rounded prematurely, thus affecting their answer to the next part.”

Worked example 13 marks

Solve 2 × 3^(x) = 50. Give your answer correct to 3 significant figures.

  1. 1

    Divide by 2: 3^(x) = 25 [1]

  2. 2

    Take logarithms of both sides: xlg 3 = lg 25, so x = (lg 25)/(lg 3) [1]

  3. 3

    x = 2.9299… = 2.93 [1]

6. Equations that become quadratics

Write every term as a power of one base. Then treat that power as one unknown, u = a^(x), and solve the quadratic equation.

a^(x) is always positive. A negative value of u gives no solution for x: reject it.

Use the laws of indices, not logarithms, on a sum or difference of terms.

a^(m+n) = a^(m) × aⁿ

a^(mn) = (a^(m))ⁿ

Examiner insight

“Some candidates used laws of indices in an invalid way.” “Many candidates rejected the negative solution from their equation.”

Worked example 14 marks

Solve the equation 2 × 8^(2x+1) - 3 × 2^(3x+4) - 64 = 0.

  1. 1

    8^(2x+1) = (2³)^(2x+1) = 2^(6x+3), so 2 × 8^(2x+1) = 2^(6x+4) = 16(2^(3x))²

  2. 2

    3 × 2^(3x+4) = 3 × 2⁴ × 2^(3x) = 48(2^(3x))

  3. 3

    16(2^(3x))² - 48(2^(3x)) - 64 = 0 [1]

  4. 4

    Divide by 16 and let u = 2^(3x): u² - 3u - 4 = 0, so (u - 4)(u + 1) = 0 [1]

  5. 5

    2^(3x) = 4 or 2^(3x) = -1. Reject -1, as 2^(3x) > 0 [1]

  6. 6

    2^(3x) = 2², so 3x = 2 and x = 2/3 [1]

7. Solving logarithmic equations

Change every log to one base. Then either combine them into a single logarithm on each side, or let y = log_a x and solve the quadratic equation in y. Rewrite log_a x = k as x = a^(k).

The value of a log can be negative. The argument of a log cannot. Check each solution in the original equation.

Examiner insight

“A few candidates made slips when rearranging to quadratic form. These candidates often misinterpreted the square as being the square of the argument of the logarithm instead of the square of the logarithm.” “Candidates should be encouraged to check that the solutions they have found to equations such as this, where the original functions are logarithms, are valid for the original form.”

Worked example 16 marks

Given that log₅(x²) - log₅ x - 3log_x 5 = 2, find x.

  1. 1

    log₅(x²) = 2log₅ x and 3log_x 5 = 3/(log₅x) [1]

  2. 2

    Let y = log₅ x: 2y - y - 3/y = 2, so y - 3/y = 2 [1]

  3. 3

    Multiply by y: y² - 2y - 3 = 0 [1]

  4. 4

    (y - 3)(y + 1) = 0 [1]

  5. 5

    log₅ x = 3 or log₅ x = -1 [1]

  6. 6

    x = 5³ = 125 or x = 5⁻¹ = 1/5. Both are valid: a log may be negative [1]

Recap

  • a^(x) = b means x = log_a b. ln x = log_e x and lg x = log₁₀ x.
  • e^(x) and ln x are each the inverse of the other. Reflect in y = x.
  • y = ke^(nx) + a has asymptote y = a. y = kln(ax + b) has asymptote x = -b/a.
  • Add logs to multiply, subtract logs to divide, and a power comes down in front.
  • Change of base: log_a b = (log_c b)/(log_c a) and log_a b = 1/(log_b a).
  • a^(x) = b: take logs of both sides. a^(x) is never negative.
  • A log can be negative. Its argument cannot. Check solutions in the original equation.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State the inverse function of f(x) = e^(x).1 mark
  2. Do not use a calculator. Write down the value of log₂ 32 and the value of lg 0.001.2 marks
  3. Do not use a calculator. State the equation of the asymptote of the curve y = 5e^(-x) + 2 and find the exact coordinates of its y-intercept.2 marks
  4. A curve has equation y = 3ln(2x - 4). Find the set of values of x for which y exists and state the equation of the asymptote.3 marks
  5. Do not use a calculator. Write 2 + lg a - 3lg b as a single logarithm to base 10.3 marks
  6. Do not use a calculator. Write 1/(log₇ e) + 2ln 3 as a single natural logarithm.3 marks
  7. Solve 3 × 4^(x) = 90. Give your answer correct to 3 significant figures.3 marks
  8. Do not use a calculator. Solve 9^(x) - 4 × 3^(x) - 45 = 0.4 marks
  9. Do not use a calculator. (a) Solve (log₃ x)² - log₉(x⁴) = 3. (b) Solve 2log₃ x = 1 + log₃(x + 6).9 marks
  10. Two metal bars are kept in different rooms. After t hours the temperature of bar A is T_A °C and the temperature of bar B is T_B °C, where T_A = 480e^(-2t) + 20 and T_B = 30 - 20e^(-t). (a) State the equation of the asymptote of the graph of T_A against t. (b) Find the exact time at which the temperature of bar A is 50 °C. Give your answer in the form kln 2. (c) Find the exact time at which the two bars have the same temperature, and find this temperature.10 marks

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