Exponential/logarithmic inverse relationship | Year 12

01 / LEARN

The idea, step by step

A logarithm asks for an exponent

For $b > 0$ and $b \neq 1$ , the statement $y = {log}_{b} x$ means $b^{y} = x$ . It asks which exponent of the fixed base produces the input. Real powers of a positive base are positive, so a real logarithm requires $x > 0$ . A negative logarithm value is allowed: it describes an input between zero and one when $b > 1$ .

y = {log}_{b} x x = b^{y} (b > 0, b \neq 1, x > 0)

Inverse functions swap input and output

The exponential function $f (x) = b^{x}$ has domain all real numbers and range $y > 0$ . Its inverse $f^{- 1} (x) = {log}_{b} x$ has domain $x > 0$ and range all real numbers. Points swap coordinates, and the graphs reflect in $y = x$ . Always retain the original domain when cancelling a logarithm with its base.

A common trap

A negative logarithm value means that its input must be negative.

Instead: ${log}_{2} (1 / 8) = - 3$ : the input is positive, while the exponent is negative. A real logarithm cannot take a nonpositive input.

02 / WORKED EXAMPLES

See the thinking

Example 1

Read a negative exponent

Evaluate ${log}_{4} (1 / 8)$ .

Write $4 = 2^{2}$ and $1 / 8 = 2^{- 3}$ .
If $4^{y} = 1 / 8$ , then $2^{2 y} = 2^{- 3}$ .
Equate exponents: $2 y = - 3$ .

- 3 / 2

Example 2

Find an inverse with a shifted input

Find the inverse of $f (x) = 3^{x - 2}$ and state both domains.

Set $y = 3^{x - 2}$ and take base-3 logarithms: ${log}_{3} y = x - 2$ .
Solve for the old input: $x = 2 + {log}_{3} y$ .
Swap variable names. The original output is positive, so the inverse input must be positive.

f^{- 1} (x) = 2 + {log}_{3} x

x > 0

;

f

has domain

ℝ

Check your understanding

If ${log}_{5} a = - 2$ , what is $a$ ?

Show answer

$a = 1 / 25$

Translate directly: $a = 5^{- 2} = 1 / 25$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Evaluate ${log}_{3} 81$ .
2 marks
Evaluate ${log}_{2} (1 / 16)$ .
2 marks
Write $7^{- 1} = 1 / 7$ as a logarithmic statement.
2 marks
Solve ${log}_{4} x = 3 / 2$ .
2 marks
Find the real domain of $g (x) = {log}_{2} (x + 5)$ .
2 marks
Find the inverse of $h (x) = 5^{2 x}$ and its domain.
3 marks

Read the worked solutions

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

Evaluate ${log}_{3} 81$ .
2 marks
$3^{4} = 81$ , so the required exponent is $4$ .
$4$
Evaluate ${log}_{2} (1 / 16)$ .
2 marks
$1 / 16 = 2^{- 4}$ .
$- 4$
Write $7^{- 1} = 1 / 7$ as a logarithmic statement.
2 marks
The logarithm records the exponent of base 7.
${log}_{7} (1 / 7) = - 1$
Solve ${log}_{4} x = 3 / 2$ .
2 marks
$x = 4^{3 / 2} = {(\sqrt{4})}^{3} = 8$ .
$x = 8$
Find the real domain of $g (x) = {log}_{2} (x + 5)$ .
2 marks
The input must satisfy $x + 5 > 0$ .
$x > - 5$
Find the inverse of $h (x) = 5^{2 x}$ and its domain.
3 marks
${log}_{5} y = 2 x$ . Solve for $x$ , then interchange names.
$h^{- 1} (x) = \frac{1}{2} {log}_{5} x$ , $x > 0$

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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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