Logarithm laws and change of base | Year 12

01 / LEARN

The idea, step by step

Why logarithm laws work

For a fixed valid base, multiplication of positive inputs becomes addition of logarithms. Division becomes subtraction, and a power becomes a multiplier. These facts follow from the corresponding index laws. They do not give a rule for the logarithm of a sum: ${log}_{b} (u + v)$ cannot generally be split.

{log}_{b} (u v) = {log}_{b} u + {log}_{b} v; {log}_{b} (u / v) = {log}_{b} u - {log}_{b} v; {log}_{b} (u^{k}) = k {log}_{b} u

Keep restrictions and change the base

Use these laws when the inputs $u$ and $v$ are positive. Simplification must not silently enlarge the original domain. To calculate an unfamiliar base, take a common logarithm or natural logarithm of $b^{y} = a$ and solve for $y$ . Use the same logarithm base in numerator and denominator.

{log}_{b} a = \frac{ln a}{ln b} (a > 0, b > 0, b \neq 1)

A common trap

$ln (x + 4) = ln x + ln 4$ .

Instead: The sum of logarithms is $ln (4 x)$ , not $ln (x + 4)$ . Test $x = 1$ : $ln 5 \neq ln 4$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Condense with restrictions

For $x > 0$ and $y > 0$ , condense $2 ln x - \frac{1}{2} ln y$ into one logarithm.

The power law gives $ln (x^{2}) - ln (\sqrt{y})$ .
Use the quotient law on the two positive inputs.

ln (\frac{x^{2}}{\sqrt{y}})

Example 2

Evaluate a nonstandard base

Calculate ${log}_{3} 7$ to three decimal places.

Use change of base: ${log}_{3} 7 = ln 7 / ln 3$ .
Calculate with unrounded stored values before rounding the final result.

1.771

Check your understanding

For $x > 0$ , simplify $ln (3 x) - ln x$ .

Show answer

$ln 3$

The quotient is $3 x / x = 3$ , on the original domain $x > 0$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $a, b > 0$ , expand $ln (5 a^{2} / b)$ .
3 marks
For $x > 0$ , condense $3 {log}_{2} x + {log}_{2} 5$ .
2 marks
Find ${log}_{5} 12$ to three decimal places.
2 marks
Simplify ${log}_{2} 48 - {log}_{2} 3$ .
2 marks
Explain why replacing $ln (x^{2})$ by $2 ln x$ requires care.
3 marks
If $ln a = 2$ and $ln b = - 1$ , find $ln (a^{2} / b^{3})$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $a, b > 0$ , expand $ln (5 a^{2} / b)$ .
3 marks
Use product, power and quotient laws; the factors are positive.
$ln 5 + 2 ln a - ln b$
For $x > 0$ , condense $3 {log}_{2} x + {log}_{2} 5$ .
2 marks
$3 {log}_{2} x = {log}_{2} (x^{3})$ , then combine the product.
${log}_{2} (5 x^{3})$
Find ${log}_{5} 12$ to three decimal places.
2 marks
${log}_{5} 12 = ln 12 / ln 5 \approx 1.543959$ .
$1.544$
Simplify ${log}_{2} 48 - {log}_{2} 3$ .
2 marks
${log}_{2} (48 / 3) = {log}_{2} 16 = 4$ .
$4$
Explain why replacing $ln (x^{2})$ by $2 ln x$ requires care.
3 marks
The original domain is $x \neq 0$ . On that domain, $ln (x^{2}) = 2 ln ∣ x ∣$ .
The replacement is valid for $x > 0$ ; $ln (x^{2})$ is also defined for $x < 0$ .
If $ln a = 2$ and $ln b = - 1$ , find $ln (a^{2} / b^{3})$ .
2 marks
$2 ln a - 3 ln b = 4 + 3 = 7$ .
$7$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: logarithmic functions and their applications; reciprocal-linear antiderivatives are taught in the Integration chapter.

Official curriculum source ↗

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