Laws of Logarithms (AA SL)
Logarithms undo exponents, and the three laws below let you combine or split them the same way index laws work for powers. Once you can move confidently between a sum of logs and a single logarithm, questions that look intimidating collapse into a couple of lines. It's part of the broader Logs & Exponents topic.
24 questions on this sub-topic.
The three laws
Covered under IB syllabus reference SL1.7: laws of logarithms, building on the introduction to logarithms in SL1.5 (\(a^x=b\) is equivalent to \(\log_a b = x\), with \(a>0\), \(b>0\), and \(\log_e x = \ln x\)). All three laws below are in the formula booklet.
Product & quotient laws
\(\log_a(xy) = \log_a x + \log_a y\)
\(\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y\)
A product inside a log becomes a sum of logs; a quotient becomes a difference. Only multiplication and division inside the log convert this way.
Power law
\(\log_a(x^m) = m\log_a x\)
An exponent inside a log can be brought out front as a multiplier - and the reverse move (coefficient in, exponent out) is just as useful for combining terms into one logarithm.
Need change of base or how to solve equations with these laws? See Exponential & Log Equations. For index laws and the full overview, see Logs & Exponents.
Worked examples
Write \(2\log_a 3 + \log_a 5 - \log_a 9\) as a single logarithm.
Worked solution
\(2\log_a 3=\log_a 3^2=\log_a 9.\) M1
\(\log_a 9+\log_a 5-\log_a 9=\log_a\!\left(\dfrac{9\times5}{9}\right).\) A1
\(=\log_a 5.\) A1
Given \(\log_a b = c\), show that \(\log_b a = \dfrac1c\) (\(a,b>0,\ a,b\ne1\)).
Worked solution
\(\log_a b=c\) means \(a^c=b.\) M1
\(\log_b a^c=\log_b b=1.\) M1
\(c\log_b a=1\Rightarrow \log_b a\) A1 \(=\dfrac1c.\) A1 AG
Given \(\log_{10}2=0.301\) and \(\log_{10}3=0.477\), find \(\log_{10}18\).
Worked solution
\(18=2\cdot3^2.\) M1
\(\log_{10}18=\log_{10}2+2\log_{10}3=0.301+2(0.477).\) A1
\(=0.301+0.954=1.255.\) A1
Solve \(\ln(x+2) + \ln(x-1) = \ln 4\).
Worked solution
The two terms add, so use the product law \(\ln A+\ln B=\ln(AB)\): \(\ln\big[(x+2)(x-1)\big]=\ln 4.\) M1
(since \(\ln\) is one-to-one): \((x+2)(x-1)=4.\) M1
\(x^2+x-2=4\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0.\) A1
Both \(\ln(x+2)\) and \(\ln(x-1)\) require \(x>1\), so reject \(x=-3.\) R1
Therefore \(x=2.\) A1
Common mistakes
- Splitting \(\log(x+y)\) into \(\log x + \log y\). Logarithms don't distribute over addition or subtraction - the product/quotient laws only apply to multiplication and division inside the log.
- Forgetting to bring a coefficient inside before combining. \(2\log_a 3\) has to become \(\log_a 9\) using the power law before it can merge with other log terms - trying to add it directly leaves the expression only half simplified.
- Losing track of which law removes which operation. The product law replaces addition, the quotient law replaces subtraction - swap them and the sign of the combined expression comes out wrong.
Ready to practise properly?
23 log-law questions, marked instantly like the real exam.
Quick answers
What are the three laws of logarithms?
The product law \(\log_a(xy) = \log_a x + \log_a y\), the quotient law \(\log_a(x/y) = \log_a x - \log_a y\), and the power law \(\log_a(x^m) = m\log_a x\).
Can I split \(\log(x+y)\) into \(\log x + \log y\)?
No. The product law only applies to multiplication and division inside the logarithm - there is no rule for splitting a sum or difference inside a log.