Sketch natural exponential and logarithmic graphs
Use inverse domains, ranges, intercepts and asymptotes, and distinguish positive, negative and zero exponential rates.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.2 and 3.2 include log laws excluding change of base, natural exponential and logarithmic graphs, equations and inequalities, and straight-line transformations. Written practice is self-checked.
Before you begin
- Understand inverse functions and reflection in y = x.
- Use logarithms and indices with positive arguments.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the domain of y = ln x over the reals?
What you will learn
- Sketch eˣ and ln x as inverse functions with correctly stated domains and ranges.
- Describe e^(kx) for positive, negative and zero k without treating an asymptote as an attained value.
The inverse functions exchange domains and ranges
The natural exponential base e is approximately 2.71828. The function y = eˣ has domain all real x, range y > 0 and intercept (0, 1). It increases and approaches the horizontal asymptote y = 0 as x tends to negative infinity, while remaining positive at every finite input. Its inverse y = ln x has domain x > 0, range all real y and intercept (1, 0). Its vertical asymptote is x = 0, approached from the positive side.
Their graphs reflect in y = x: for example (ln 2, 2) on eˣ corresponds to (2, ln 2) on ln x. The inverse identities ln(eˣ) = x for real x and e^(ln x) = x for positive x use different input conditions. A negative logarithm output is compatible with a positive argument below 1. Neither graph crosses its asymptote, and zero is not a valid input for ln x.
The sign of an exponential rate determines its direction
For y = e^(kx), every real x is allowed and the output is positive. If k > 0, the graph increases and approaches zero towards negative x; if k < 0 it decreases and approaches zero towards positive x. Both pass through (0, 1), and their range is y > 0. If k = 0, the function is the constant 1, so its range is just {1}. Reflecting eˣ in the y-axis gives e^(−x), illustrating a negative rate.
For y = e^(−2x), the graph decreases, has intercept (0, 1) and gives y = 1/2 at x = (ln 2)/2. In a supplied decay model C(t) = e^(−t/2) for t ≥ 0, the restricted range is 0 < C ≤ 1; it reaches 1/2 at t = 2 ln 2 but never reaches zero at a finite time. Restrict the model's domain separately from the all-real domain of the mathematical exponential.
Pause and explain
For y = e^(−2x), which statement is correct?
Worked example
Describe and sketch y = e^(−2x), including its domain, range, intercept and a point where y = 1/2.
Show the worked solution
- The negative coefficient gives a decreasing positive exponential with domain all real x and intercept (0, 1).
- Its range is y > 0; as x grows it approaches y = 0 without attaining it.
- Set −2x = ln(1/2) = −ln 2 to obtain x = (ln 2)/2.
Answer Decreasing; domain ℝ; range y > 0; intercept (0, 1); point ((ln 2)/2, 1/2).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe y = e^(2x) and find its point with y = 3.
Give me a hint
Use the sign of 2 and invert the equation e^(2x) = 3.
Compare my reasoning
- The graph is increasing with domain ℝ, range y > 0 and intercept (0, 1).
- Taking ln gives 2x = ln 3, hence x = (ln 3)/2.
- The point is ((ln 3)/2, 3); the graph approaches y = 0 towards negative x but does not meet it.
Increasing; domain ℝ; range y > 0; point ((ln 3)/2, 3).
Look for these in your work
- State domain, range and direction correctly.
- Use the inverse relation to locate an exact point.
Sketch y = ln x, stating its inverse, domain, range, intercept and value at x = e^(−2).
Give me a hint
Exchange the domain and range of eˣ and use ln(eᵘ) = u.
Compare my reasoning
- The inverse is eˣ; ln x has domain x > 0 and range ℝ.
- It crosses at (1, 0) and has vertical asymptote x = 0 approached from the right.
- At positive input e^(−2), its value is −2, demonstrating that valid log outputs can be negative.
Inverse eˣ; domain x > 0; range ℝ; intercept (1, 0); ln(e^(−2)) = −2.
Look for these in your work
- Keep the positive input domain separate from the all-real output range.
- Label the intercept and correct side of the asymptote.
A supplied normalised signal model is C(t) = e^(−t/2) for t ≥ 0. Find its initial value, time when C = 1/2 and range on the stated domain. Can it reach zero at a finite time in this model?
Give me a hint
Use the restricted time domain and invert the half-value equation exactly.
Compare my reasoning
- C(0) = 1, and the negative rate makes the model decrease for t ≥ 0.
- From −t/2 = ln(1/2) = −ln 2, the half-value time is t = 2 ln 2.
- The restricted range is 0 < C ≤ 1; no finite t gives zero because an exponential remains positive.
Initial value 1; half-value time 2 ln 2; range 0 < C ≤ 1; no finite zero.
Look for these in your work
- Apply the model's non-negative time domain.
- Distinguish an asymptotic limit from an attained output.
Common mistakes
- An exponential approaches zero without taking value zero at a finite real input.
- The domain of ln x is positive x, while its output may be negative.
- A model's restricted domain changes its range; k = 0 gives a constant rather than growth or decay.
What can you explain now?
Evaluate ln(e³), e^(ln 4) and ln(e^(−2)).
Compare with the explanation
3, 4 and −2 respectively.
Use the inverse identities; the last logarithm has positive argument e^(−2) and a negative output.
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Make it stick
Sketch eˣ, ln x and e^(−x) with their key boundaries. Explain why an inverse can have negative outputs while its input domain stays positive.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.