Natural logarithms | Year 12

01 / LEARN

The idea, step by step

Why base e is useful

The number $e$ is approximately $2.71828$ . The natural logarithm $ln x$ means ${log}_{e} x$ , not a new operation unrelated to logarithms. It is especially useful in calculus because the derivative of $e^{x}$ is itself. Inverse cancellation gives $ln (e^{u}) = u$ for any real $u$ , and $e^{ln v} = v$ only for $v > 0$ .

ln e = 1, ln 1 = 0, ln (e^{u}) = u

Solving a natural-log equation

Isolate the logarithm, then apply the exponential function to both sides. If logarithms are combined or the equation is squared, check every candidate in the original expression. The positivity restriction concerns each logarithm input, not the final logarithm value.

A common trap

$ln (e^{2} + e^{3}) = 2 + 3$ .

Instead: Cancellation applies to one exponential input, not a sum. Factor first: $ln (e^{2} (1 + e)) = 2 + ln (1 + e)$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Keep an exact value

Simplify $ln (e^{- 3}) + ln (e^{2})$ .

Inverse cancellation gives the separate values $- 3$ and $2$ .
Add these values, without rounding $e$ .

- 1

Example 2

Check original inputs

Solve $ln (x - 1) + ln (x + 1) = ln 8$ .

The original domain is $x > 1$ .
Combine the logs: $ln (x^{2} - 1) = ln 8$ , so $x^{2} - 1 = 8$ .
The algebraic candidates are $x = 3, - 3$ . Only $x = 3$ satisfies $x > 1$ .

x = 3

Check your understanding

Solve $ln (2 x - 1) = 0$ .

Show answer

$x = 1$

$2 x - 1 = e^{0} = 1$ , and $x = 1$ is in the domain $x > 1 / 2$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $ln (e^{7 / 2})$ .
2 marks
Simplify $e^{2 ln 5}$ .
2 marks
Solve $ln (x + 4) = 2$ .
2 marks
Solve $2 ln x = ln (3 x + 4)$ .
3 marks
Order $ln (1 / 2)$ , $0$ , $ln 2$ and $1$ from least to greatest.
2 marks
Simplify $ln (2 e^{3})$ exactly.
2 marks

Read the worked solutions

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

Evaluate $ln (e^{7 / 2})$ .
2 marks
Natural logarithm and exponential are inverse operations.
$7 / 2$
Simplify $e^{2 ln 5}$ .
2 marks
$2 ln 5 = ln 25$ and $e^{ln 25} = 25$ .
$25$
Solve $ln (x + 4) = 2$ .
2 marks
$x + 4 = e^{2}$ . The resulting input is positive.
$x = e^{2} - 4$
Solve $2 ln x = ln (3 x + 4)$ .
3 marks
Domain $x > 0$ . Combine to $ln (x^{2}) = ln (3 x + 4)$ . Then $(x - 4) (x + 1) = 0$ ; reject $- 1$ .
$x = 4$
Order $ln (1 / 2)$ , $0$ , $ln 2$ and $1$ from least to greatest.
2 marks
$1 / 2 < 1 < 2 < e$ , and $ln x$ is increasing.
$ln (1 / 2) < 0 < ln 2 < 1$
Simplify $ln (2 e^{3})$ exactly.
2 marks
Use the product law and $ln (e^{3}) = 3$ .
$ln 2 + 3$

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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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