Pearson Edexcel International Advanced Level Mathematics: Exponentials, logs and numerical methods (YMA01)

This guide teaches P3 sections 3 and 6 (topics 3.1 to 3.3 and 6.1 to 6.2) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that P3 is compulsory for IAL Mathematics and Pure Mathematics, and that its IAS weighting is N/A, so it is not part of IAS. It also says P1 and P2 knowledge is assumed, including the laws of logarithms, so revisit the Unit P2 overview if those feel rusty.

Pair it with the revision notes for recall and the practice questions for testing. Course links: the Edexcel A Level Mathematics hub, the printable checklist and all free 10-minute diagnostics.

What the sections cover

Topic	What you must be able to do
3.1	Know the function eˣ and its graph, including graphs of the form y = e^(ax + b) + c
3.2	Know ln x and its graph, and that ln x is the inverse function of eˣ; solve e^(ax + b) = p and ln(ax + b) = q
3.3	Estimate the constants in y = axⁿ and y = kbˣ from straight-line logarithmic graphs
6.1	Locate roots of f(x) = 0 by looking for a change of sign in an interval where f(x) is continuous
6.2	Find approximate solutions with simple iterative methods, including xₙ₊₁ = f(xₙ)

3.1 The function eˣ and its graph

The number e is about 2.718. The graph of y = eˣ has four features to learn.

• It passes through (0, 1).
• It is always positive, so the range is y > 0.
• The x-axis (y = 0) is a horizontal asymptote on the left: as x → −∞, eˣ → 0.
• It is increasing everywhere, and it grows faster than any power of x as x → ∞.

The family y = e^(ax + b) + c. Read it as a sequence of changes to y = eˣ. The constant c lifts the whole curve, so the asymptote moves from y = 0 to y = c and the range becomes y > c. The constants a and b change the horizontal position and scale but do not change the asymptote. If a is negative, the curve is a mirror image and decreases.

Worked example 1. Sketch the key features of y = e^(3x + 2) − 4.

Asymptote: y = −4, and the range is y > −4.
y-intercept (x = 0): y = e² − 4 = 3.389 (3 d.p.)
x-intercept (y = 0): e^(3x + 2) = 4
                     3x + 2 = ln 4
                     x = (ln 4 − 2)/3 = −0.205 (3 d.p.)
Shape: increasing, because a = 3 is positive.

Mark the asymptote as a dashed line and label both intercepts. A curve that crosses its own asymptote, or that dips below y = −4, is wrong.

3.2 The function ln x and its graph

The natural logarithm ln x is the inverse function of eˣ. That gives two cancelling rules, valid wherever both sides exist:

e^(ln x) = x   (x > 0 only)
ln(eˣ) = x       (any real x)

Inverse graphs are reflections in the line y = x, so reflect y = eˣ in that line to get y = ln x. The new curve passes through (1, 0), approaches the y-axis (x = 0) without touching it, exists only for x > 0, and takes every real value of y. The point (a, b) on y = eˣ maps to (b, a) on y = ln x.

Solving e^(ax + b) = p. Take ln of both sides, then rearrange. This needs p > 0.

Solving ln(ax + b) = q. Write both sides as powers of e, so ax + b = e^q. This is positive automatically; the domain check matters when you combine logarithms first.

Worked example 2. Solve (a) e^(5 − 2x) = 12, (b) e^(2x − 5) = 40, giving exact answers and then 3 significant figures.

(a) 5 − 2x = ln 12
    x = (5 − ln 12)/2 = 1.26

(b) 2x − 5 = ln 40
    x = (5 + ln 40)/2 = 4.34

Worked example 3. Solve (a) ln(3x + 4) = 2, (b) ln(5 − 2x) = −1.

(a) 3x + 4 = e²
    x = (e² − 4)/3 = 1.13

(b) 5 − 2x = e⁻¹
    x = (5 − e⁻¹)/2 = 2.32

In (b) the argument 5 − 2x equals e⁻¹, which is positive, so the answer is valid; substituting x = 2.32 back is a sensible check. Leave answers exact (such as (e² − 4)/3) unless the question asks for a decimal.

3.3 Logarithmic graphs

Some relationships become straight lines when you take logarithms of the data. Use either lg (base 10) or ln, but use the same base throughout.

The power law y = axⁿ. Take logs of both sides:

log y = n log x + log a

Compare with a line Y = mX + c, where Y is log y and X is log x. Plot log y against log x. The gradient is n, and the intercept on the vertical axis is log a.

The exponential law y = kbˣ. Take logs again:

log y = (log b) x + log k

Plot log y against x. The gradient is log b, and the intercept is log k.

Which graph? If a straight line appears when you plot log y against log x, the data fit a power law. If it appears against x, they fit an exponential law. The first form needs both variables logged; the second logs only y.

Worked example 4 (power law). Readings of x and y are modelled by y = axⁿ. Values of lg x and lg y are:

x	1	2	3	4	5	6
lg x	0	0.301	0.477	0.602	0.699	0.778
lg y	0.398	0.820	1.064	1.241	1.377	1.487

The points lie on a straight line. Use the first and last points to find a and n.

gradient = (1.487 − 0.398)/(0.778 − 0) = 1.40   so n = 1.40
intercept on the lg y axis = 0.398
lg a = 0.398   so a = 10^0.398 = 2.50
Model: y = 2.50x^1.4

Using two points that are far apart gives a more reliable gradient than using two that are close together.

Worked example 5 (exponential law). A mass of y grams of a substance is measured x days after the start. Values of lg y are:

x	0	1	2	3	4	5
lg y	1.903	1.833	1.762	1.691	1.621	1.550

The graph of lg y against x is a straight line. Find k and b in y = kbˣ.

intercept = 1.903        lg k = 1.903    k = 10^1.903 = 80.0
gradient = (1.550 − 1.903)/5 = −0.0706   lg b = −0.0706
b = 10^(−0.0706) = 0.850
Model: y = 80 × 0.85ˣ

A negative gradient means b < 1, so the quantity decays. The value k is the value of y when x = 0.

6.1 Locating roots by sign change

Suppose f(x) is continuous on [a, b]. If f(a) < 0 < f(b), or f(b) < 0 < f(a), then f(x) = 0 has at least one root in that interval. Continuous means the graph can be drawn without lifting the pen.

Worked example 6. Show that e^x = 5x has a root in [2.5, 2.6].

f(x) = eˣ − 5x
f(2.5) = −0.3175 (negative)
f(2.6) = +0.4637 (positive)
f is continuous and changes sign across [2.5, 2.6], so it has a root there.

Always state the values, the sign change and the continuity. To justify a root to 2 d.p., find the interval of width 0.01 either side. Here f(2.54) = −0.0203 and f(2.55) = 0.0571, so the root is 2.54 to 2 d.p.

When a sign change misleads.

f(x) = 1/(x − 2):  f(1) = −1, f(3) = 1, but there is no root.
The function is not continuous at x = 2.

When no sign change hides a root. For f(x) = (x − 1)² there is a root at x = 1 but f(x) ≥ 0 on both sides, so no sign change appears. Two roots inside one interval also give the same sign at both ends. So the absence of a sign change proves nothing: for f(x) = eˣ − 5x, f(0) = 1 and f(3) = 5.09 are both positive, although the interval [0, 3] contains two roots.

6.2 Approximate solutions by iteration

An iterative formula xₙ₊₁ = f(xₙ) generates a sequence from a starting value x₀. If the sequence settles on a number, that number α satisfies α = f(α), so it is a root of the rearranged equation. The guidance for 6.2 says that leads will be given, so a formula is normally supplied.

Worked example 7. The equation eˣ − 5x = 0 can be rearranged to x = ln(5x). Use xₙ₊₁ = ln(5xₙ) with x₀ = 2.5.

x₁ = ln 12.5      = 2.5257
x₂ = ln 12.6287   = 2.5360
x₃ = 2.5400,  x₄ = 2.5416,  x₅ = 2.5422,  x₆ = 2.5425

The differences shrink, and the values approach 2.543. To confirm 2.543 to 3 d.p., test f(2.5425) = −0.0011 and f(2.5435) = 0.0066, a sign change.

Convergence depends on the formula and the start. The same equation also rearranges to x = eˣ/5. Starting at x₀ = 0.3 the terms are 0.2700, 0.2620, 0.2599, 0.2594 and settle near 0.2592, the smaller root. Starting at x₀ = 2.6 the terms are 2.69, 2.95, 3.84, 9.29 and grow without limit, so that formula cannot find the larger root.

A staircase diagram shows this: draw y = f(x) and y = x, step vertically to the curve, then horizontally to the line. If the curve is shallow at the root, the steps close in; if it is steep, they move away.

Common errors

• Writing ln(x + y) as ln x + ln y, or e^(a + b) as e^a + e^b.
• Taking ln of both sides of an equation but forgetting to rearrange the remaining linear equation.
• Accepting a solution of an equation involving ln(ax + b), after combining logarithms, without checking that ax + b is positive.
• Reading the gradient of a log graph as a, or the intercept as n; for y = axⁿ the gradient is n and the intercept is log a.
• Plotting log y against x for a power law, or log y against log x for an exponential law.
• Rounding in the middle of iteration; keep at least two more decimal places than you need.
• Claiming a root from a sign change without saying f is continuous.

Next, try the practice set, and keep the revision notes for last-minute recall.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Unit P3, sections 3 and 6 (Pearson Education Limited).