Logarithm laws | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

Products become sums, quotients become differences

Logarithm laws translate index laws into operations on logarithms. For a common valid base and positive arguments, a product becomes a sum and a quotient becomes a difference. A power on a positive argument becomes a coefficient outside the logarithm.

Keep brackets around whole arguments. The power law applies to ${log}_{b} (x^{k})$ , while the product law applies to ${log}_{b} (x y)$ . Neither law splits a sum inside an argument.

Write the original domain first

Write the original domain before combining logarithms. Each original argument must be positive on its own. A combined product or quotient may also be positive at inputs that were never allowed in the original expression.

After solving the transformed equation, test every candidate against those original restrictions and substitute valid ones back. This final step removes extraneous roots without discarding legitimate solutions.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Expand a product and quotient

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

Worked solution

1
Separate the quotient and product.
${log}_{2} (\frac{8 x^{3}}{y}) = {log}_{2} 8 + {log}_{2} (x^{3}) - {log}_{2} y$
The supplied conditions $x, y > 0$ make all separated logarithms valid.
2
Bring the power outside.
${log}_{2} 8 + 3 {log}_{2} x - {log}_{2} y$
The exponent on a positive argument becomes a multiplier of its logarithm.
3
Evaluate the constant logarithm.
$3 + 3 {log}_{2} x - {log}_{2} y$
Since $2^{3} = 8$ , the constant term is ${log}_{2} 8 = 3$ .

Answer

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

EXAMPLE 02

Reject an extraneous root

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

Worked solution

1
State the original domain.
$x > 0, x - 3 > 0 ⟹ x > 3$
Both original logarithm arguments must be positive.
2
Combine the logarithms and exponentiate.
$log (x (x - 3)) = 1 ⟹ x (x - 3) = 10$
The logs use base ten, so an argument with logarithm 1 equals 10.
3
Solve the resulting quadratic.
$x^{2} - 3 x - 10 = (x - 5) (x + 2) = 0 ⟹ x = 5 or x = - 2$
Factoring gives every algebraic candidate.
4
Apply the domain and verify.
$x = 5 : log 5 + log 2 = log 10 = 1$
The root $- 2$ violates $x > 3$ ; 5 satisfies the original equation.

Answer

x=5

03 / YOUR TURN

Try it yourself.

Check your understanding

Combine $log 6 + log 5$ .

Show answer & reasoning

$log 30$

The arguments are positive and the bases match, so the product law gives $log 6 + log 5 = log (6 \cdot 5) = log 30$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Combine $log a + log b$ for $a, b > 0$ .
2 marks
Expand $log (\frac{100 x^{2}}{y})$ 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 $log (\frac{100 x^{2}}{y})$ 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.
$log (\frac{x^{2}}{y})$
Solve $log (x + 4) - log x = log 3$ .
3 marks
Domain $x > 0$ ; $\frac{x + 4}{x} = 3$ , giving $x + 4 = 3 x$ 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 ↗

In this lesson

Understand the idea.

Products become sums, quotients become differences

Write the original domain first

See the method in action.

Expand a product and quotient

Separate the quotient and product.

Bring the power outside.

Evaluate the constant logarithm.

Reject an extraneous root

State the original domain.

Combine the logarithms and exponentiate.

Solve the resulting quadratic.

Apply the domain and verify.

Try it yourself.

Check your understanding

Keep practising