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Logarithms

The logarithm <b>laws</b> are senior content. The Year 10 requirement is logarithmic <b>scales</b> (AC9M10M02), covered in the Measurement guide.

A logarithm asks one question: what power do I raise the base to, to get this number?

Builds onYr 8 · Exponent laws

Method

The definition
log⁡ab=x  ⟺  ax=b\log_{a} b = x \iff a^{x} = b
A log is an exponent. That is the whole idea.
Log of the base
log⁡aa=1\log_{a} a = 1
Because $a^{1} = a$.
Log of one
log⁡a1=0\log_{a} 1 = 0
Because $a^{0} = 1$.
Product
log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log(ab)
Adding logs multiplies the numbers.
Quotient
log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log\left(\tfrac{a}{b}\right)
Subtracting logs divides them.
Power
log⁡an=nlog⁡a\log a^{n} = n \log a
The exponent comes out the front.

Worked example

e.g. log⁡232\log_{2} 32
  1. Ask the question the log is asking: 2 to what power gives 32?2?=322^{?} = 32
  2. Write 32 as a power of the base32=2532 = 2^{5}
  3. The answer is that exponentlog⁡232=5\log_{2} 32 = 5
  4. Check by going back the other way25=32 ✓2^{5} = 32 \ \checkmark

Common mistakes

Wronglog⁡a+log⁡b=log⁡(a+b)\log a + \log b = \log(a + b)
Rightlog⁡a+log⁡b=log⁡(ab)\log a + \log b = \log(ab)
Adding logs multiplies the numbers, because logs are exponents and exponents add when you multiply.
Wronglog⁡20=0\log_{2} 0 = 0
Rightlog⁡20 does not exist\log_{2} 0 \text{ does not exist}
No power of 2 gives 0. You can never take the log of zero or a negative number.
Wronglog⁡alog⁡b=log⁡a−log⁡b\frac{\log a}{\log b} = \log a - \log b
Rightlog⁡a−log⁡b=log⁡(ab)\log a - \log b = \log\left(\tfrac{a}{b}\right)
Dividing two logs is not the same as subtracting them. Only the subtraction rule turns into a fraction inside the log.

Practice

Evaluate:

1
log⁡28\log_{2} 8
Answer33
2
log⁡381\log_{3} 81
Answer44
3
log⁡525\log_{5} 25
Answer22
4
log⁡101000\log_{10} 1000
Answer33
5
log⁡77\log_{7} 7
Answer11
6
log⁡41\log_{4} 1
Answer00
7
log⁡264\log_{2} 64
Answer66
8
log⁡10100+log⁡1010\log_{10} 100 + \log_{10} 10
Answer33
9
log⁡216−log⁡24\log_{2} 16 - \log_{2} 4
Answer22
10
log⁡392\log_{3} 9^{2}
Answer44
11
log⁡218\log_{2} \tfrac{1}{8}
Answer−3-3
12
Solve log⁡2x=5\text{Solve } \log_{2} x = 5
Answerx=32x = 32
Next step
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