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Logarithmic and Exponential Functions

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  Cambridge International AS & A Level Mathematics 9709 — Pure Mathematics 3

Logarithmic and Exponential Functions

The big idea: A logarithm is an index: loga y = x means ax = y. Taking logs brings an unknown down from the power, and turns curved relationships into straight lines.

Summary — What This Chapter Covers
  • loga y = x ⇔ ax = y. The laws: log(xy) = log x + log y, log(x/y) = log x − log y, log xk = k log x.
  • ln x = loge x. ex and ln x are inverse functions: eln x = x and ln(ex) = x.
  • y = ekx is growth for k > 0 and decay for k < 0; both pass through (0, 1).
  • Take logs of both sides to solve equations or inequalities with the unknown in an index.
  • y = kxn becomes ln y = ln k + n ln x; y = k(ax) becomes ln y = ln k + x ln a.
1. Laws of Logarithms and ex
Logs and e

Indices written another way

Every law of logarithms is a law of indices in disguise. The laws hold for any base, including e. Change of base is not required for this paper.

The graph of y = ex passes through (0, 1), is always positive and has the x-axis as an asymptote. y = ln x is its reflection in y = x: it passes through (1, 0), is defined only for x > 0 and has the y-axis as an asymptote. For y = ekx: k > 0 rises steeply; k < 0 falls towards the x-axis.

Laws log(xy) = log x + log y    log(x/y) = log x − log y    log xk = k log x    eln x = x    ln ex = x
Worked Example

Solve ln(2x + 3) = 2 ln x + ln 3, giving x to 3 significant figures.

Practice Question

Solve e2x − 5ex + 6 = 0.

2. Solving Equations and Linear Forms
Logs in action

Unknown indices and straight-line graphs

To solve af(x) = bg(x), take ln of both sides: f(x) ln a = g(x) ln b, a linear equation when f and g are linear. For an inequality, the same step is safe because ln is an increasing function, but dividing by a negative log (such as ln 0.5) reverses the sign.

To find constants in a model, take ln of both sides to get the form Y = mX + c, then read the gradient and intercept from the straight-line graph.

Linear forms y = kxn ⇒ ln y = n ln x + ln k    (plot ln y against ln x: gradient n, intercept ln k)
y = k(ax) ⇒ ln y = x ln a + ln k    (plot ln y against x: gradient ln a, intercept ln k)
Worked Example 1

Solve 3x + 1 = 4x − 1, giving x to 3 significant figures.

Worked Example 2

Variables x and y satisfy y = kxn. The graph of ln y against ln x is a straight line through (0.5, 2.1) and (2.0, 5.1). Find n and k.

Practice Question 1

3x − 1

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