OCR A Level Mathematics A: Exponential and logarithmic functions: e^x, a^x, ln x and the gradient of e^kx (H240) – Revision Notes

These are condensed notes for OCR A Level Mathematics A (H240), based on the OCR specification (Version 3.1, August 2026): content references 1.06a to 1.06e. All of this is Stage 1 content, common to AS Level (H230) and A Level. Laws of logarithms, exponential equations a^x = b, linear form and modelling (1.06f to 1.06i) are separate topics. Full explanations are in the study guide; test yourself with the practice set. See also the course hub, the checklist and the free diagnostics.

Key results

Ref	Result
1.06a	y = a^x (a > 0): passes through (0, 1), always positive, asymptote y = 0
1.06a	e ≈ 2.718; y = e^x is increasing with asymptote y = 0 and gradient 1 at (0, 1)
1.06b	gradient of e^(kx) is k e^(kx)
1.06c	y = a^x ⇔ x = log_a y; log_a a = 1; log_a 1 = 0
1.06d	ln x = log_e x; domain x > 0; passes through (1, 0) and (e, 1)
1.06e	ln(e^x) = x for all x; e^(ln x) = x for x > 0; ln e = 1; ln 1 = 0

1.06a Graphs of a^x and e^x

• a > 1: increasing. 0 < a < 1: decreasing. a = 1: the line y = 1.
• Every one passes through (0, 1) and never touches the x-axis.
• y = a^(−x) is y = (1/a)^x, the reflection of y = a^x in the y-axis.
• Adding a constant moves the asymptote: y = a^x + c has asymptote y = c and y-intercept 1 + c.
• Sequences link: terms of C × r^t form a geometric sequence with ratio r.

y = 2^x:  x   -2    -1    0   1   2   3
          y   1/4   1/2   1   2   4   8

When two models are compared, tabulate both over a sensible range and look for the crossover. A larger base eventually wins, however small its starting value.

1.06b Gradient of e^(kx)

Method in steps:

1. Write the function as a constant times e^(kx), identifying k.
2. Multiply by k: the gradient function is k × the original.
3. Substitute the given x to get the numerical gradient.
4. For a tangent, find y at that x, then use y − y₁ = m(x − x₁).
5. For a normal, use the negative reciprocal of m.

Why it matters: the gradient is k times the value, so the rate of change is proportional to the current size. A positive k gives growth, a negative k decay.

Reminder. For y = e^(3x) at x = 0: y = 1, gradient 3, tangent y = 3x + 1.

1.06c Logarithms

A logarithm is a power: log_a b is the power of a that gives b. Convert both ways:

5⁻² = 1/25   ⇔   log₅ (1/25) = -2
log₂ 32 = 5  ⇔   2⁵ = 32

Method for an unknown in a log equation:

1. Rewrite as a power using x = log_a y ⇔ y = a^x.
2. Solve the resulting equation by indices.
3. Check that the base is positive and any bracket inside the log is positive.

Graph of y = log_a x (a > 1): through (1, 0) and (a, 1), asymptote x = 0, domain x > 0, range all reals. It is the reflection of y = a^x in y = x.

Replacing x by x − c in a log graph moves the asymptote to x = c and the domain to x > c.

1.06d and 1.06e ln x and e^x

ln is the logarithm to base e. It undoes e^x, and e^x undoes ln.

Method for e^(f(x)) = k:

1. Take ln of both sides: f(x) = ln k (needs k > 0).
2. Rearrange for x.

Method for ln(f(x)) = k:

1. Raise e to both sides: f(x) = e^k.
2. Rearrange for x.

Method for the inverse of an e^x or ln x function:

1. Write y = f(x) and rearrange to get e^(…) or ln(…) on its own.
2. Apply the other function to both sides.
3. Make x the subject, then swap the letters.
4. State the domain as the range of f.

Reminder. If f(x) = e^x + 2 then f(x) > 2, f⁻¹(x) = ln(x − 2) and f⁻¹ accepts only x > 2.

Worked reminders

Gradient condition. Find the point on y = 4e^(x/2) where the gradient is 6.

dy/dx = 4 × (1/2) e^(x/2) = 2e^(x/2) = 6   →   e^(x/2) = 3
x/2 = ln 3   →   x = 2 ln 3 = 2.20
y = 4 × 3 = 12

The point is (2 ln 3, 12). The y-coordinate comes from e^(ln 3) = 3.

Rate of change. A quantity is Q = 5e^(−0.2t). Then dQ/dt = −e^(−0.2t) = −0.2Q. At t = 5, Q = 5e^(−1) = 1.84 and the rate is −e^(−1) = −0.368. The negative sign shows decay, and the rate is always the same fraction, −0.2, of the current value.

Log graph. For y = log₃ x, the points (1/3, −1), (1, 0), (3, 1) and (9, 2) lie on the curve. Reflect them in y = x to get points of y = 3^x: (−1, 1/3), (0, 1), (1, 3), (2, 9).

Sketching checklist

Before you sketch an exponential or logarithmic graph, check these in order:

1. Is the base greater than 1 or between 0 and 1?
2. Mark the asymptote as a dashed line and label its equation.
3. Mark the intercept with the y-axis (for a^x types) or x-axis (for log types).
4. Mark one more labelled point so the scale is clear.
5. Show the curve approaching the asymptote without touching it.

Values to know

Expression	Value	Reason
log_a a	1	a¹ = a
log_a 1	0	a⁰ = 1
ln e	1	e¹ = e
ln 1	0	e⁰ = 1
e⁰	1	any non-zero base to the power 0
e^(ln 5)	5	inverse functions
ln(e^(−3))	−3	inverse functions

Must-know distinctions

Compare	Key difference
a^x and x^a	the variable is in the power for a^x; a^x has a horizontal asymptote and is positive
log_a x and a^x	inverses: swap the coordinates of every point; the asymptote switches from y = 0 to x = 0
e^(kx) and e^x	the gradient of e^(kx) is k times the value; for e^x the gradient equals the value
ln x and log x	ln is base e; always say which base
f⁻¹ and 1/f	f⁻¹ is the inverse function; 1/f is the reciprocal
growth and decay	k > 0 gives growth, k < 0 gives decay; both stay positive

Quick self-test

1. State the y-intercept and asymptote of y = 5^x + 2.
2. Is y = 0.4^x increasing or decreasing? How is its graph related to y = 2.5^x?
3. Find the gradient of y = e^(−2x) at x = 1, to 3 significant figures.
4. Find the gradient of y = 3e^(x/3) at x = 3.
5. Write 5⁻² = 1/25 in logarithmic form.
6. Evaluate log₉ 3.
7. Solve ln(x + 1) = 2, giving an exact answer and a 3 significant figure answer.
8. Solve 2e^(3x) = 14, giving an exact answer.
9. State the domain and range of y = ln x.
10. Find the inverse of f(x) = e^x + 2 and state its domain.

Answers

1. (0, 3); y = 2.
2. Decreasing, since 0 < 0.4 < 1. 0.4^x = (2.5)^(−x), so it is the reflection of y = 2.5^x in the y-axis.
3. Gradient = −2e^(−2) = −0.271.
4. Gradient = 3 × (1/3) e^(x/3) = e^(x/3); at x = 3 it is e = 2.72.
5. log₅ (1/25) = −2.
6. 9^(1/2) = 3, so log₉ 3 = 1/2.
7. x + 1 = e², so x = e² − 1 = 6.39.
8. e^(3x) = 7, so 3x = ln 7 and x = (ln 7)/3 = 0.649.
9. Domain x > 0; range all real numbers.
10. f⁻¹(x) = ln(x − 2), domain x > 2.

Where marks are usually lost

• Drawing a^x with the asymptote at the wrong axis, or crossing it.
• Forgetting that the asymptote of a^x + c is y = c, not y = 0.
• Differentiating e^(kx) without the factor k, or putting k in the power only.
• Taking the tangent gradient from the wrong x, or using the normal’s gradient as the tangent’s.
• Writing log_a b as a/b or as a number times b.
• Leaving an answer as a decimal when the question asks for an exact value such as ln 7 or e² − 1.
• Rounding an intermediate value and losing accuracy in the final answer.
• Taking ln of a negative number or zero, for instance when solving e^x = −3.
• Giving an inverse function with no domain, or with the domain of the original function.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: section 1.06, Exponentials and Logarithms, references 1.06a to 1.06e.