Use logarithm laws with valid arguments and bases
Connect logarithms with indices, combine products and quotients correctly and preserve the domain of every original logarithm.
Section 2.2 · Logarithmic and exponential functions
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
- Use index laws, including fractional and negative powers.
- Solve simple polynomial equations and check sign restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does log₂(1/4) equal?
What you will learn
- Interpret logₐ x = p as aᵖ = x with a > 0, a ≠ 1 and x > 0.
- Apply product, quotient and power laws while checking original arguments are positive.
A logarithm names the exponent of a positive base
For a > 0 with a ≠ 1, logₐ x = p means aᵖ = x, requiring x > 0 for a real logarithm. Thus log₂ 8 = 3 and log₂(1/4) = −2. A negative logarithm value is allowed; a negative argument is not. The base 1 is excluded because 1ᵖ does not determine a unique exponent. In ln x the base is the natural exponential constant e.
With a valid common base and positive x and y, logₐ(xy) = logₐ x + logₐ y, logₐ(x/y) = logₐ x − logₐ y, and logₐ(xʳ) = r logₐ x. There is no matching law for a sum: ln(x + y) does not generally equal ln x + ln y. The stated positive-argument conditions matter when applying a power law; a rewritten expression must keep the original permitted inputs. Change of base is outside this section's required log-laws content.
Combine the expression, then recheck the original domains
For positive x and y, ln(8x³/√y) = ln 8 + 3 ln x − (1/2)ln y = 3 ln 2 + 3 ln x − (1/2)ln y. The coefficient of a logarithm records an exponent, and a division becomes a difference. Keep one common base throughout a calculation rather than combine logarithms with different bases as if they were the same quantity.
For ln(x − 1) + ln(x + 1) = ln 8, the original arguments require x > 1. Combining gives ln(x² − 1) = ln 8, hence x² = 9. The algebraic candidates are ±3, but only x = 3 satisfies the original domain. The combined argument x² − 1 alone would also be positive at −3; checking it alone loses the restrictions of the two original logarithms.
Pause and explain
The combined equation ln(x² − 1) = ln 8 gives ±3. Which solves ln(x − 1) + ln(x + 1) = ln 8?
Worked example
For x > 0 and y > 0, expand ln(8x³/√y) using log laws.
Show the worked solution
- Use the quotient law to separate ln(8x³) − ln(√y).
- Use product and power laws to obtain ln 8 + 3 ln x − (1/2)ln y.
- Since 8 = 2³, ln 8 = 3 ln 2; retain the stated positive x and y domain.
Answer 3 ln 2 + 3 ln x − (1/2)ln y, for x > 0 and y > 0.
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.
For x > 0 and y > 0, expand log₂(8x²/y).
Give me a hint
Separate product and quotient terms, then use the known power 8 = 2³.
Compare my reasoning
- The quotient law gives log₂(8x²) − log₂ y.
- The product and power laws give log₂ 8 + 2 log₂ x − log₂ y.
- Since log₂ 8 = 3, the result is 3 + 2 log₂ x − log₂ y on the stated domain.
3 + 2 log₂ x − log₂ y, for x > 0 and y > 0.
Look for these in your work
- Use product, quotient and power laws with one base.
- Retain the positive-argument domain.
Solve ln(x − 1) + ln(x + 1) = ln 8 and explain any rejected root.
Give me a hint
State the domain before combining the logarithms.
Compare my reasoning
- The original arguments require x − 1 > 0 and x + 1 > 0, hence x > 1.
- Combine to ln(x² − 1) = ln 8; equality of logs gives x² = 9 and candidates ±3.
- At x = 3 the arguments 2 and 4 are valid and multiply to 8; reject −3 because its original arguments are negative.
x = 3 only; x = −3 fails the original log domains.
Look for these in your work
- Check each original argument, not just the combined product.
- Verify the retained root in the original equation.
Positive dimensionless measurements R and S satisfy ln R − 2 ln S = ln 5. Find R when S = 2 and explain the recovered relationship.
Give me a hint
Use the power and quotient laws to combine the left side before removing the logarithm.
Compare my reasoning
- For positive R and S, ln R − 2 ln S = ln(R/S²).
- Equality with ln 5 gives R/S² = 5, so R = 5S².
- At S = 2, R = 20; substituting gives ln 20 − 2 ln 2 = ln 5.
R = 5S²; when S = 2, R = 20.
Look for these in your work
- Interpret the coefficient 2 as a power.
- Check the positive values in the original log relationship.
Common mistakes
- A log law for a product is not a log law for a sum.
- Keep a common valid base and positive arguments.
- After combining logs, check each original argument before retaining equation roots.
What can you explain now?
Simplify ln 3 + ln 4, and explain why the answer is not ln 7.
Compare with the explanation
ln 12, not ln 7.
Adding logarithms combines a product, 3 × 4, rather than the sum of the two arguments.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the base and argument conditions for a real logarithm. Give a product-law calculation and show why original log domains can reject a root after combining.
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.