Logarithms
Understand logarithms for Queensland Year 11 Mathematical Methods (QCAA). A logarithm is really an index: it answers what power a base must be raised to, so it reverses raising a base to a power.
You will learn to define logarithms as indices, convert between exponential and logarithmic forms, evaluate logarithms to a base greater than one, and apply the logarithmic laws — groundwork for solving exponential equations and modelling change.
Every question with a fully worked solution.
- Logarithms - Video - Logarithms Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), a logarithm answers the question “to what power must the base be raised?”: \(a^{y}=x\) is the same statement as \(y=\log_a x\). This page defines logarithms to a base \(a>1\), evaluates them by switching to index form, and applies the logarithmic laws (product, quotient and power) to combine, expand and simplify logarithm expressions.
A logarithm is an index (a power). For a base \(a>1\) and \(x>0\), the logarithm form \(y=\log_a x\) means exactly the same as the index form \(a^{y}=x\). So \(\log_a x\) is “the power that \(a\) must be raised to in order to give \(x\)”.
Two special values follow straight from the definition: \(\log_a a=1\) (because \(a^{1}=a\)) and \(\log_a 1=0\) (because \(a^{0}=1\)). The argument \(x\) must be positive — there is no logarithm of \(0\) or of a negative number.
The logarithmic laws come from the index laws: adding logarithms multiplies their arguments, subtracting divides them, and a power on the argument comes out the front as a multiplier.
Definition (for \(a>1,\ x>0\)):
The logarithmic laws (each on its own line):
How to evaluate or simplify a logarithm
- Identify the base \(a\) and the argument, and check the argument is positive.
- Apply the laws one at a time — combine sums/differences into a single logarithm, and bring any power to the front — showing one law per line.
- Convert to index form \(a^{y}=x\), write the argument as a power of \(a\), and equate the indices to read off the value.
Set the logarithm equal to \(x\) and write it in index form:
| \(\log_3 81\) | \(=\) | \(x\) |
| \(3^{x}\) | \(=\) | \(81\) |
Write \(81\) as a power of \(3\), then equate the indices:
| \(3^{x}\) | \(=\) | \(3^{4}\) |
| \(x\) | \(=\) | \(4\) |
\(\log_3 81 = 4\).
Quotient law — a subtraction of logs becomes a division:
| \(\log_2 48-\log_2 3\) | \(=\) | \(\log_2\!\left(\dfrac{48}{3}\right)\) |
| \(=\) | \(\log_2 16\) |
Convert to index form and equate indices:
| \(\log_2 16\) | \(=\) | \(x\) |
| \(2^{x}\) | \(=\) | \(16=2^{4}\) |
| \(x\) | \(=\) | \(4\) |
\(\log_2 48-\log_2 3 = 4\).
Power law on the first term — the coefficient becomes an index:
| \(2\log_{10}5\) | \(=\) | \(\log_{10}5^{2}\) |
| \(=\) | \(\log_{10}25\) |
Product law (the sum) then quotient law (the difference):
| \(\log_{10}25+\log_{10}8-\log_{10}2\) | \(=\) | \(\log_{10}\!\left(\dfrac{25\times 8}{2}\right)\) |
| \(=\) | \(\log_{10}100\) |
Convert to index form and equate indices:
| \(\log_{10}100\) | \(=\) | \(y\) |
| \(10^{y}\) | \(=\) | \(100=10^{2}\) |
| \(y\) | \(=\) | \(2\) |
\(2\log_{10}5+\log_{10}8-\log_{10}2 = \log_{10}100 = 2\).
Product law — combine the two logs on the left:
| \(\log_4\big(x(x-6)\big)\) | \(=\) | \(2\) |
Rewrite in index form (this is the equate-the-indices step in reverse):
| \(x(x-6)\) | \(=\) | \(4^{2}\) |
| \(x^{2}-6x\) | \(=\) | \(16\) |
Solve the quadratic:
| \(x^{2}-6x-16\) | \(=\) | \(0\) |
| \((x-8)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(8\ \text{ or }\ x=-2\) |
Reject \(x=-2\): the argument of a logarithm must be positive (and \(x-6\) would be negative). Check \(x=8\): \(\log_4 8+\log_4 2=\log_4 16=2\). \(\checkmark\)
\(x=8\).
Common pitfalls
Frequently asked questions
What is a logarithm in simple terms?
It is a power. \(\log_a x\) is the power you must raise the base \(a\) to in order to get \(x\); that is, \(a^{y}=x\) means \(y=\log_a x\).
How do you evaluate a logarithm like log base 2 of 32?
Set it equal to \(y\), write \(2^{y}=32\), then \(32=2^{5}\), so equating indices gives \(y=5\).
What are the three logarithm laws?
\(\log_a x+\log_a y=\log_a(xy)\); \(\log_a x-\log_a y=\log_a\!\left(\dfrac{x}{y}\right)\); and \(\log_a(x^{n})=n\log_a x\).
Why is log base a of 1 always zero?
Because \(a^{0}=1\) for any base \(a>1\). By the definition of a logarithm, \(\log_a 1=0\).
Can you take the logarithm of a negative number in Year 11?
No. For a base \(a>1\) the argument must be positive, so \(\log_a x\) is only defined for \(x>0\).
Do these pages use natural logarithms or e?
No. Year 11 Methods uses logarithms to a base \(a>1\) (often base \(10\)); natural logarithms and \(e\) are Year 12.