Logarithm laws
Logarithm laws change products, quotients and powers into simpler equivalent expressions. The base must be valid and every logarithm argument must be positive
\[\log_{a}\left(xy\right)=\log_{a}\left(x\right)+\log_{a}\left(y\right)\quad \log_{a}\left(\frac{x}{y}\right)=\log_{a}\left(x\right)-\log_{a}\left(y\right)\]
- Use the product law The logarithm of a product is the sum of the logarithms of its positive factors \(\displaystyle \log_{a}\left(xy\right)=\log_{a}\left(x\right)+\log_{a}\left(y\right)\)
- Use the quotient law The logarithm of a quotient is the difference between the logarithms of its positive parts \(\displaystyle \log_{a}\left(\frac{x}{y}\right)=\log_{a}\left(x\right)-\log_{a}\left(y\right)\)
- Use the power law An exponent on a positive logarithm argument becomes a coefficient outside the logarithm \(\displaystyle \log_{a}\left(x^{r}\right)=r\log_{a}\left(x\right)\)
- Check inverse conditions Inverse cancellation is valid when the base is positive and not 1 and the logarithm argument is positive \(\displaystyle \log_{a}\left(a^{r}\right)=r,\quad a>0,\quad a\ne 1\)
Worked example
Use logarithm laws to find the exact value of the expression below
\[\log_{3}\left(27\right)+\log_{3}\left(9\right)\]
- \(\displaystyle 27=3^{3}\quad\text{and}\quad 9=3^{2}\)Rewrite each positive argument as a power of the common base
- \(\displaystyle \log_{3}\left(27\right)+\log_{3}\left(9\right)=\log_{3}\left(27\times 9\right)=\log_{3}\left(243\right)\)The sum of logarithms with the same base condenses to the logarithm of a product
- \(\displaystyle 243=3^{5}\quad\Longrightarrow\quad \log_{3}\left(243\right)=5\)The exponent that produces 243 from base 3 is 5
Common mistakes
Splitting a logarithm of a sum
✕\(\displaystyle \log_{2}\left(8+4\right)=\log_{2}\left(8\right)+\log_{2}\left(4\right)\)✓\(\displaystyle \log_{2}\left(8\times 4\right)=\log_{2}\left(8\right)+\log_{2}\left(4\right)\)
The product law applies to multiplication inside the logarithm, not to additionMultiplying logarithm values
✕\(\displaystyle \log_{2}\left(8\times 4\right)=\log_{2}\left(8\right)\times\log_{2}\left(4\right)=6\)✓\(\displaystyle \log_{2}\left(8\times 4\right)=\log_{2}\left(8\right)+\log_{2}\left(4\right)=5\)
A product inside a logarithm becomes a sum of logarithms, not a product of their valuesDividing logarithms for a quotient
✕\(\displaystyle \log_{2}\left(\frac{8}{4}\right)=\frac{\log_{2}\left(8\right)}{\log_{2}\left(4\right)}=\frac{3}{2}\)✓\(\displaystyle \log_{2}\left(\frac{8}{4}\right)=\log_{2}\left(8\right)-\log_{2}\left(4\right)=1\)
A quotient inside a logarithm becomes a difference of logarithms, not a quotient of their valuesDifficulty levels
introOne product or quotient law is used with an exact evaluation
standardA power law is combined with one product or quotient law
challengeSeveral laws are combined or an inverse identity follows an index simplification
examA four-option expression choice is paired with a table of coefficients and argument exponents
Questions about this topic
Can a logarithm of a sum be split into two logarithms?+
No. The product and quotient laws apply to multiplication and division, not to addition or subtraction inside a logarithm
Why does a quotient become a difference?+
The quotient law follows from subtracting exponents when dividing powers with the same base
When can inverse logarithm identities be used?+
They can be used when the base is positive and not 1 and the logarithm argument is positive
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 11: Exponentials and logarithms
In this chapter
Simplify and evaluate powersApply logarithm laws and inverse identitiesConnect exponential and logarithmic formsSolve exponential equations and inequalities exactlyConstruct exponential and logarithmic inverse pairsSolve exponential and logarithmic equationsAnalyse exponential and logarithmic graphsInterpret and use an exponential model