Logarithm laws | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

Products become sums, quotients become differences

For positive x and y and a valid common base b, ${log}_{b} (x y) = {log}_{b} x + {log}_{b} y$ and ${log}_{b} (x / y) = {log}_{b} x - {log}_{b} y$ . These follow from index laws.

A power can move in front: ${log}_{b} (x^{k}) = k {log}_{b} x$ for x>0. Use brackets to make the argument clear. All logs combined by one law must have the same base.

Write the original domain first

Combining logs may disguise their original restrictions. If a starting expression contains both log(x) and log(x-3), its real domain is x>3, even if the combined product happens to be positive elsewhere.

There is no sum law that splits $log (x + y)$ into two logs. When solving a log equation, obtain candidate roots algebraically and substitute them back into the original logarithmic equation.

A common trap

Using $log (x + y) = log x + log y$ .

Instead: The addition law belongs to a product inside the logarithm, not a sum.

02 / WORKED EXAMPLES

See the thinking

Example 1

Expand a product and quotient

Expand ${log}_{2} (8 x^{3} / y)$ for x,y>0.

Use product, quotient and power laws.
${log}_{2} 8 + 3 {log}_{2} x - {log}_{2} y = 3 + 3 {log}_{2} x - {log}_{2} y$ .

3 + 3 {log}_{2} x - {log}_{2} y

Example 2

Reject an extraneous root

Solve $log x + log (x - 3) = 1$ .

The original domain is x>3.
Combine and exponentiate: $x (x - 3) = 10$ .
$x^{2} - 3 x - 10 = (x - 5) (x + 2) = 0$ . Candidates are 5 and -2.
Only x=5 belongs to the original domain; $log 5 + log 2 = log 10 = 1$ .

x=5

Check your understanding

Combine $log 6 + log 5$ .

Show answer

$log 30$

Both arguments are positive and the bases match.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Combine $log a + log b$ for a,b>0.
2 marks
Expand $log (100 x^{2} / y)$ for x,y>0.
2 marks
Combine $2 log x - log y$ for x,y>0.
2 marks
Solve $log (x + 4) - log x = log 3$ .
3 marks
Solve $log x + log (x - 2) = log 8$ .
3 marks
Compare $log (1 + 9)$ with $log 1 + log 9$ . Explain what this tests.
3 marks

Read the worked solutions

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

Combine $log a + log b$ for a,b>0.
2 marks
A product inside one log equals the sum of the logs.
$log (a b)$
Expand $log (100 x^{2} / y)$ for x,y>0.
2 marks
$log 100 = 2$ ; apply quotient and power laws.
$2 + 2 log x - log y$
Combine $2 log x - log y$ for x,y>0.
2 marks
Move 2 into the exponent, then use the quotient law.
$log (x^{2} / y)$
Solve $log (x + 4) - log x = log 3$ .
3 marks
Domain x>0; $(x + 4) / x = 3$ , giving x+4=3x and x=2.
x=2
Solve $log x + log (x - 2) = log 8$ .
3 marks
Domain x>2; $x (x - 2) = 8$ , so $(x - 4) (x + 2) = 0$ .
Reject -2; retain 4.
x=4
Compare $log (1 + 9)$ with $log 1 + log 9$ . Explain what this tests.
3 marks
$log 10 = 1$ but $log 1 + log 9 = 0 + log 9$ .
The first is 1; the second is approximately 0.95424; the proposed sum-splitting rule fails.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

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