Logarithms and the logarithm laws
In Year 12 Mathematical Methods (Queensland, QCAA), a logarithm is an index: \(a^{x}=b\) is equivalent to \(x=\log_{a}b\) (for \(a>0,\ a\neq 1\)). This page covers evaluating logarithms, the logarithm laws — product, quotient, power, change of base, with \(\log_{a}a=1\) and \(\log_{a}1=0\) — and using them to simplify expressions and solve equations, the algebra behind the calculus of \(\ln x\) in Unit 3.
A logarithm is the index (power) a base must be raised to. The logarithm base \(a\) of \(b\), written \(\log_{a}b\), is defined by
\(a^{x}=b \quad\Longleftrightarrow\quad x=\log_{a}b\), where \(a>0,\ a\neq 1\) and \(b>0\).
So \(\log_{2}8=3\) because \(2^{3}=8\), and \(\log_{10}1000=3\) because \(10^{3}=1000\). Two special values follow straight from the definition: \(\log_{a}a=1\) (since \(a^{1}=a\)) and \(\log_{a}1=0\) (since \(a^{0}=1\)). The natural logarithm \(\ln x=\log_{e}x\) uses base \(e\approx 2.718\), and \(\log x\) with no base written means base \(10\).
Because logarithms are indices, the index laws give matching logarithm laws. A product becomes a sum, a quotient becomes a difference, and a power becomes a multiplier:
- Product: \(\log_{a}(xy)=\log_{a}x+\log_{a}y\)
- Quotient: \(\log_{a}\!\left(\dfrac{x}{y}\right)=\log_{a}x-\log_{a}y\)
- Power: \(\log_{a}(x^{n})=n\log_{a}x\)
The change of base rule \(\log_{a}x=\dfrac{\log_{b}x}{\log_{b}a}\) lets you evaluate any base on a calculator (using base \(10\) or base \(e\)), and solve exponential equations such as \(2^{x}=10\).
The definition of a logarithm (base \(a>0,\ a\neq 1\)):
The logarithm laws (product, quotient, power):
The change of base rule (evaluate any base with a calculator):
How to simplify a logarithmic expression
- Powers first. Use the power law to move any coefficient inside as an index: \(2\log_{3}6=\log_{3}6^{2}=\log_{3}36\).
- Combine. Use the product law for a sum and the quotient law for a difference, into a single logarithm: \(\log_{3}36-\log_{3}4=\log_{3}\dfrac{36}{4}=\log_{3}9\).
- Evaluate. Write the argument as a power of the base and read off the index: \(\log_{3}9=\log_{3}3^{2}=2\).
(a) \(32=2^{5}\), so \(\log_{2}32=5\).
(b) \(1000=10^{3}\), so \(\log_{10}1000=3\).
(c) \(\ln e=\log_{e}e=1\) (since \(e^{1}=e\)).
The point \((8,3)\) is on the curve, so \(\log_{2}8=3\) (check: \(2^{3}=8\)).
It cuts the \(x\)-axis where \(y=0\), i.e. at \((1,0)\), since \(\log_{2}1=0\).
(a) Product law, then evaluate.
| \(\log_{6}4+\log_{6}9\) | \(=\) | \(\log_{6}(4\times 9)\) |
| \(=\) | \(\log_{6}36=\log_{6}6^{2}=2\) |
(b) Quotient law, then evaluate.
| \(\log_{2}40-\log_{2}5\) | \(=\) | \(\log_{2}\dfrac{40}{5}\) |
| \(=\) | \(\log_{2}8=\log_{2}2^{3}=3\) |
Combine with the product law, then rewrite in index form.
| \(\log_{2}\big(x(x-2)\big)\) | \(=\) | \(3\) |
| \(x(x-2)\) | \(=\) | \(2^{3}=8\) |
| \(x^{2}-2x-8\) | \(=\) | \(0\) |
| \((x-4)(x+2)\) | \(=\) | \(0\) |
So \(x=4\) or \(x=-2\). Reject \(x=-2\) (it makes \(\log_{2}x\) undefined). \(x=4\).
Common pitfalls
Frequently asked questions
What is a logarithm?
A logarithm is an index. \(\log_{a}b\) is the power of \(a\) that gives \(b\): \(a^{x}=b\Leftrightarrow x=\log_{a}b\), for \(a>0,\ a\neq 1\).
How do you evaluate a logarithm such as log base 2 of 8?
Write the number as a power of the base: \(8=2^{3}\), so \(\log_{2}8=3\). Also \(\log_{a}a=1\) and \(\log_{a}1=0\).
What are the logarithm laws?
Product \(\log_{a}(xy)=\log_{a}x+\log_{a}y\); quotient \(\log_{a}\dfrac{x}{y}=\log_{a}x-\log_{a}y\); power \(\log_{a}(x^{n})=n\log_{a}x\).
What is the change of base rule?
\(\log_{a}x=\dfrac{\log_{b}x}{\log_{b}a}\). For example \(\log_{2}10=\dfrac{\log 10}{\log 2}\approx 3.32\).
How do you solve a logarithmic equation using the laws?
Combine to one logarithm, rewrite in index form, solve, then check every solution keeps each argument positive.
Why is log base a of (x plus y) not log base a of x plus log base a of y?
The product law applies to a product, not a sum: \(\log_{a}x+\log_{a}y=\log_{a}(xy)\). There is no law for the log of a sum.