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.
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.