Logarithm laws and change of base | Year 12

01 / LEARN

Understand the idea.

Why logarithm laws work

Logarithm laws come from index laws: multiplying positive inputs adds their exponents, dividing subtracts them, and raising a positive input to a power multiplies its logarithm by that power. Use one fixed base with $b > 0$ , $b \neq 1$ .

These laws concern products, quotients and powers. There is no general splitting rule for ${log}_{b} (u + v)$ . Before expanding or condensing an expression, identify its actual operation and ensure each original logarithm input is positive.

\begin{aligned} {log}_{b} (u v) & = {log}_{b} u + {log}_{b} v \\ {log}_{b} (\frac{u}{v}) & = {log}_{b} u - {log}_{b} v \\ {log}_{b} (u^{k}) & = k {log}_{b} u \end{aligned}

Keep restrictions and change the base

Condensing can hide restrictions. The expression $ln (x - 1) + ln (x + 1)$ requires both inputs positive, even though the single expression $ln (x^{2} - 1)$ also exists for some negative values of $x$ . Retain the original domain.

For an unfamiliar base, write $b^{y} = a$ and take natural logarithms: $y ln b = ln a$ . Thus change of base divides logarithms of the input and base, calculated using the same logarithm function without early rounding.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Condense with restrictions

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

Worked solution

1
Move the coefficients into powers.
$2 ln x - \frac{1}{2} ln y = ln (x^{2}) - ln (y^{1 / 2})$
The power law is valid because the stated x and y are positive.
2
Combine the difference as a quotient.
$ln (x^{2}) - ln (\sqrt{y}) = ln (\frac{x^{2}}{\sqrt{y}})$
Subtracting logarithms divides their positive inputs.
3
Retain the original restrictions.
$ln (\frac{x^{2}}{\sqrt{y}}), x > 0, y > 0$
A condensed expression must keep the conditions under which the original logarithms exist.

Answer

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

EXAMPLE 02

Evaluate a nonstandard base

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

Worked solution

1
Express the requested value as an exponent.
$y = {log}_{3} 7 ⟺ 3^{y} = 7$
This gives an equation whose exponent can be isolated using logarithms.
2
Take natural logarithms and solve.
$y ln 3 = ln 7 ⟹ y = \frac{ln 7}{ln 3}$
The power law brings y outside the logarithm, and ln 3 is nonzero.
3
Calculate and round once.
$\frac{ln 7}{ln 3} = 1.771243 \dots \approx 1.771$
The fourth decimal digit is 2, so three decimal places remain 1.771.

Answer

$1.771$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$ln 3$

For $x > 0$ , both original inputs are positive. The quotient law gives $ln (3 x) - ln x = ln (\frac{3 x}{x}) = ln 3$ , with the original restriction retained.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

For $a, b > 0$ , expand $ln (\frac{5 a^{2}}{b})$ .
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 $ln (\frac{a^{2}}{b^{3}})$ .
2 marks

Read the worked solutions

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

For $a, b > 0$ , expand $ln (\frac{5 a^{2}}{b})$ .
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
${log}_{5} 12 = \frac{ln 12}{ln 5} \approx 1.543959$ .
$1.544$
Simplify ${log}_{2} 48 - {log}_{2} 3$ .
2 marks
${log}_{2} (\frac{48}{3}) = {log}_{2} 16 = 4$ .
$4$
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 $ln (\frac{a^{2}}{b^{3}})$ .
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 ↗

In this lesson

Understand the idea.

Why logarithm laws work

Keep restrictions and change the base

See the method in action.

Condense with restrictions

Move the coefficients into powers.

Combine the difference as a quotient.

Retain the original restrictions.

Evaluate a nonstandard base

Express the requested value as an exponent.

Take natural logarithms and solve.

Calculate and round once.

Try it yourself.

Check your understanding

Keep practising