Natural logarithms | Year 12

01 / LEARN

Understand the idea.

Why base e is useful

The natural logarithm $ln x$ is the logarithm with base $e \approx 2.71828$ . It returns the exponent needed to produce $x$ from $e$ , which makes it especially useful for the exponential functions used in calculus.

Inverse cancellation gives $ln (e^{u}) = u$ for every real $u$ , because $e^{u} > 0$ . Conversely, $e^{ln v} = v$ requires $v > 0$ . In particular, $ln 1 = 0$ and $ln e = 1$ exactly; no decimal approximation is needed.

\begin{aligned} ln e & = 1 \\ ln 1 & = 0 \\ ln (e^{u}) & = u \end{aligned}

Solving a natural-log equation

Solve a natural-log equation by first recording all positive-input restrictions. Isolate or validly combine the logarithms, then exponentiate both sides. The exponential and logarithm reverse each other on their allowed domains.

Combining logarithms may produce an algebraic equation with candidates outside the original domain. Check each candidate against every original logarithm input. A negative logarithm output is allowed; a nonpositive input is not.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Keep an exact value

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

Worked solution

1
Cancel each inverse pair.
$ln (e^{- 3}) = - 3, ln (e^{2}) = 2$
Both exponential inputs are positive, so natural-log cancellation applies exactly.
2
Add the exact values.
$ln (e^{- 3}) + ln (e^{2}) = - 3 + 2 = - 1$
Using the inverse relationship avoids approximation of e.

Answer

$- 1$

EXAMPLE 02

Check original inputs

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

Worked solution

1
Find the original domain.
$x - 1 > 0, x + 1 > 0 ⟹ x > 1$
Both separate logarithms must exist.
2
Combine logarithms and equate their inputs.
$ln ((x - 1) (x + 1)) = ln 8 ⟹ x^{2} - 1 = 8$
The product law is valid on x>1, and ln is one-to-one on positive inputs.
3
Solve the algebraic equation.
$x^{2} = 9 ⟹ x = 3 or x = - 3$
Both square roots are algebraic candidates.
4
Check the original inputs and keep the valid root.
$x = 3 : (x - 1, x + 1) = (2, 4), ln 2 + ln 4 = ln 8$
The root 3 is valid; at -3 both inputs are negative, so that candidate is rejected.

Answer

$x = 3$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = 1$

Exponentiate $ln (2 x - 1) = 0$ to obtain $2 x - 1 = e^{0} = 1$ , so $x = 1$ . This input satisfies the original domain $x > \frac{1}{2}$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Evaluate $ln (e^{\frac{7}{2}})$ .
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 $ln (\frac{1}{2})$ , $0$ , $ln 2$ 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 $ln (e^{\frac{7}{2}})$ .
2 marks
Natural logarithm and exponential are inverse operations.
$\frac{7}{2}$
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 $ln (\frac{1}{2})$ , $0$ , $ln 2$ and $1$ from least to greatest.
2 marks
$\frac{1}{2} < 1 < 2 < e$ , and $ln x$ is increasing.
$ln (\frac{1}{2}) < 0 < ln 2 < 1$
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 ↗

In this lesson

Understand the idea.

Why base e is useful

Solving a natural-log equation

See the method in action.

Keep an exact value

Cancel each inverse pair.

Add the exact values.

Check original inputs

Find the original domain.

Combine logarithms and equate their inputs.

Solve the algebraic equation.

Check the original inputs and keep the valid root.

Try it yourself.

Check your understanding

Keep practising