Solve exponential inequalities with the correct direction
Use monotonicity, keep exact boundaries and reverse a comparison when dividing by a negative logarithmic coefficient.
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
- Solve linear and quadratic inequalities with correct endpoints.
- Apply logarithmic power laws and recognise increasing or decreasing exponentials.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When dividing an inequality by ln(1/2), what happens to its direction?
What you will learn
- Solve inequalities with unknowns in an exponent and retain strict or inclusive boundaries.
- Use the sign of a base logarithm or a decay coefficient to determine the correct inequality direction.
Taking ln preserves order, but division can reverse it
The natural logarithm is increasing on positive inputs, so taking ln preserves an inequality between two positive sides. The next step must respect the coefficient's sign. For a > 1, ln a > 0, while for 0 < a < 1, ln a < 0. Thus 2ˣ < 5 gives x < (ln 5)/(ln 2), but (1/2)ˣ > 1/8 gives x < 3 because the base is decreasing. A base of 1 is constant and must be checked directly rather than divided by ln 1 = 0.
For 3·2^(3x − 1) < 5, divide the positive coefficient 3 to get 2^(3x − 1) < 5/3. Taking ln gives (3x − 1)ln 2 < ln(5/3). Since ln 2 > 0, the solution is x < [1 + ln(5/3)/ln 2]/3. The exact boundary is approximately 0.579; it is excluded. Use the exact expression, not a rounded boundary, when checking nearby inputs.
Solve the exponent comparison and apply the model domain
Since eˣ is increasing, e^(x² − 3x) < e⁴ is equivalent to x² − 3x < 4. Factor x² − 3x − 4 = (x + 1)(x − 4), so −1 < x < 4 with both equality endpoints excluded. The exponential step can expose a polynomial inequality rather than only a linear one; finish the underlying inequality before returning to the original statement.
For a supplied decay amount q(t) = 8e^(−t/4) with t ≥ 0, the requirement q ≤ 2 becomes −t/4 ≤ ln(1/4) = −ln 4. Multiplying by −4 reverses the inequality, giving t ≥ 4 ln 4, approximately 5.55 in the model's time units. The endpoint is included because ‘at most’ allows equality. Check this interval against the original time domain and the original positive amount.
Pause and explain
For 8e^(−t/4) ≤ 2, which interval is correct on t ≥ 0?
Worked example
Solve 3·2^(3x − 1) < 5 with an exact boundary.
Show the worked solution
- Divide by 3 and take ln: (3x − 1)ln 2 < ln(5/3).
- Since ln 2 > 0, divide without reversing to get 3x − 1 < ln(5/3)/ln 2.
- Therefore x < [1 + ln(5/3)/ln 2]/3; equality at the boundary gives 5 and is excluded.
Answer x < [1 + ln(5/3)/ln 2]/3; boundary approximately 0.579, excluded.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Solve 2ˣ < 5 exactly and state whether the boundary is included.
Give me a hint
Take ln and note that ln 2 is positive.
Compare my reasoning
- The increasing ln function gives x ln 2 < ln 5.
- Divide by positive ln 2 to obtain x < (ln 5)/(ln 2).
- At the boundary 2ˣ = 5, so the strict inequality excludes that input.
x < (ln 5)/(ln 2); boundary excluded.
Look for these in your work
- Use the positive sign of ln 2.
- Keep the exact boundary and its strict endpoint treatment.
Solve (1/2)^(2x − 1) ≤ 1/8 exactly.
Give me a hint
Express 1/8 as (1/2)³ and remember that this positive base is below 1.
Compare my reasoning
- The equation compares (1/2)^(2x − 1) with (1/2)³.
- Since the base is decreasing, the condition is 2x − 1 ≥ 3.
- Hence x ≥ 2, with equality retained because both sides equal 1/8 there.
x ≥ 2, including the boundary.
Look for these in your work
- Reverse the exponent comparison for a base between 0 and 1.
- Retain the inclusive endpoint.
A supplied amount model is q(t) = 8e^(−t/4) units for t ≥ 0, with t in minutes. Find when the amount is at most 2 units, with an exact boundary.
Give me a hint
Divide by 8, take ln and track the negative coefficient of time.
Compare my reasoning
- The condition is e^(−t/4) ≤ 1/4, so −t/4 ≤ −ln 4.
- Multiplying by −4 reverses the comparison: t ≥ 4 ln 4.
- This threshold is approximately 5.55 minutes and lies in the stated time domain; equality gives exactly 2 units.
t ≥ 4 ln 4 minutes; threshold approximately 5.55 minutes, included.
Look for these in your work
- Reverse the comparison at the negative multiplier.
- Check the model domain and the inclusive amount threshold.
Common mistakes
- Reverse the comparison when dividing or multiplying by a negative coefficient.
- Do not replace an exact solution boundary with its rounded approximation.
- Check equality treatment and any stated model domain.
What can you explain now?
Solve e^(x² − 3x) < e⁴.
Compare with the explanation
−1 < x < 4.
The increasing exponential gives (x + 1)(x − 4) < 0, which holds strictly between its two roots.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain separately why taking ln preserves order and why dividing by a negative log coefficient reverses it. Check a strict and an inclusive exponential boundary.
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.