Logarithm laws | Year 10 | Ascent Library

01 / LEARN

The idea, step by step

Laws inherited from exponents

For positive $M, N$ and a valid fixed base $b$ , ${log}_{b} (M N) = {log}_{b} M + {log}_{b} N$ , ${log}_{b} (M / N) = {log}_{b} M - {log}_{b} N$ , and ${log}_{b} (M^{k}) = k {log}_{b} M$ .

The product law works because multiplying equal-base powers adds their exponents. There is no corresponding law that splits the log of a sum.

Check each original input

Combining logarithms can produce an equation with extra algebraic solutions. A result is valid only if every logarithm in the original equation has a positive input.

The change-of-base identity ${log}_{b} M = {log}_{a} M / {log}_{a} b$ lets you use a calculator with a different log base. Both bases must be positive and not equal to $1$ , and $M > 0$ .

A common trap

$log (M + N) = log M + log N$ .

Instead: The right-hand side is $log (M N)$ , not the log of a sum. For example, ${log}_{10} (10 + 10)$ is not $1 + 1$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Combine logs

Simplify ${log}_{2} 12 - {log}_{2} 3$ .

Use the quotient law: ${log}_{2} (12 / 3) = {log}_{2} 4$ .
Since $2^{2} = 4$ , the answer is $2$ .

Example 2

Solve and reject an invalid value

Solve ${log}_{10} x + {log}_{10} (x - 9) = 1$ .

The original domain is $x > 9$ .
Combine to get ${log}_{10} [x (x - 9)] = 1$ , hence $x (x - 9) = 10$ .
$x^{2} - 9 x - 10 = (x - 10) (x + 1) = 0$ . Only $x = 10$ lies in the domain.

x = 10

Check your understanding

Simplify $2 {log}_{3} 5 - {log}_{3} 25$ .

Show answer

$2 {log}_{3} 5 = {log}_{3} 25$ , so the difference is zero.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Simplify ${log}_{2} 8 + {log}_{2} 4$ .
2 marks
Write ${log}_{5} 75 - {log}_{5} 3$ as one log and evaluate.
2 marks
Expand ${log}_{10} (x^{3} / y)$ for $x, y > 0$ .
2 marks
Solve ${log}_{2} x + {log}_{2} 4 = 5$ .
2 marks
Solve ${log}_{10} (x - 1) + {log}_{10} (x + 1) = {log}_{10} 8$ .
3 marks
A student claims ${log}_{10} 2 + {log}_{10} 3 = {log}_{10} 5$ . Give the correct single log.
3 marks

Read the worked solutions

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

Simplify ${log}_{2} 8 + {log}_{2} 4$ .
2 marks
It equals ${log}_{2} 32 = 5$ .
5
Write ${log}_{5} 75 - {log}_{5} 3$ as one log and evaluate.
2 marks
${log}_{5} (75 / 3) = {log}_{5} 25 = 2$ .
2
Expand ${log}_{10} (x^{3} / y)$ for $x, y > 0$ .
2 marks
Use quotient and power laws.
$3 {log}_{10} x - {log}_{10} y$
Solve ${log}_{2} x + {log}_{2} 4 = 5$ .
2 marks
${log}_{2} (4 x) = 5$ , so $4 x = 32$ and $x = 8$ .
$x = 8$
Solve ${log}_{10} (x - 1) + {log}_{10} (x + 1) = {log}_{10} 8$ .
3 marks
Domain $x > 1$ . The product gives $x^{2} - 1 = 8$ , so $x = \pm 3$ ; only $3$ is valid.
$x = 3$
A student claims ${log}_{10} 2 + {log}_{10} 3 = {log}_{10} 5$ . Give the correct single log.
3 marks
The product law multiplies $2$ and $3$ , rather than adding them.
${log}_{10} 6$

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Integer and fractional index laws; connecting fractional powers with roots; simplifying surds; operations with surds; rationalising denominators; exact and approximate values; exponential tables and graphs; growth/decay factors; comparing linear/quadratic/exponential patterns; exponential equations; logarithms as inverses; logarithm laws; interpreting logarithmic scales; graph transformations and parameter experiments; choosing and evaluating models.

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