Exponential/logarithmic inverse relationship | Year 12

01 / LEARN

Understand the idea.

A logarithm asks for an exponent

A logarithm gives an exponent: ${log}_{b} x = y$ means $b^{y} = x$ . A valid real base has $b > 0$ and $b \neq 1$ . Since powers of a positive base are positive, the input must satisfy $x > 0$ .

The output may still be negative. For a base greater than $1$ , a negative exponent produces a positive fraction below $1$ ; for example, $2^{- 3} = \frac{1}{8}$ , so ${log}_{2} \frac{1}{8} = - 3$ .

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

Inverse functions swap input and output

The inverse of $b^{x}$ reverses its input-output relationship. Exponential inputs can be any real number and outputs are positive; logarithmic inputs are positive and outputs can be any real number. The graphs reflect in $y = x$ .

To find an inverse, write $y = f (x)$ , solve for the old input $x$ , then rename the variables. The inverse domain comes from the original range; inverse cancellation must retain any original input restrictions.

Graphs of y equals natural logarithm x and y equals exponential x on equal-unit axes, reflected in the dashed line y equals x; the points (1,0) and (0,1) are labelled.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read a negative exponent

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

Worked solution

1
Translate the logarithm into an exponent.
$y = {log}_{4} \frac{1}{8} ⟺ 4^{y} = \frac{1}{8}$
The logarithm asks which power of 4 produces one eighth.
2
Express both sides using base 2.
${(2^{2})}^{y} = 2^{- 3} ⟹ 2^{2 y} = 2^{- 3}$
A common base makes the exponents comparable.
3
Equate exponents and solve.
$2 y = - 3 ⟹ y = - \frac{3}{2}$
The base-2 exponential is one-to-one, so equal outputs have equal exponents.

Answer

$- 3 / 2$

EXAMPLE 02

Find an inverse with a shifted input

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

Worked solution

1
Write the function and its domain and range.
$y = 3^{x - 2}, x \in ℝ, y > 0$
A shifted exponent remains defined for all real inputs and has exactly the positive outputs.
2
Take a base-3 logarithm and solve for x.
${log}_{3} y = x - 2 ⟹ x = 2 + {log}_{3} y$
The logarithm reverses exponentiation with the same base.
3
Rename the inverse variables and state its domain.
$f^{- 1} (x) = 2 + {log}_{3} x, x > 0$
The inverse input is the original positive output; the original function domain is all real numbers.

Answer

$f^{- 1} (x) = 2 + {log}_{3} x$ , $x > 0$ ; $f$ has domain $ℝ$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$a = 1 / 25$

Translate the statement into exponential form: $a = 5^{- 2} = \frac{1}{5^{2}} = \frac{1}{25}$ . The negative logarithm value specifies a reciprocal power, not a negative input.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Evaluate ${log}_{3} 81$ .
2 marks
Evaluate ${log}_{2} (\frac{1}{16})$ .
2 marks
Write $7^{- 1} = \frac{1}{7}$ as a logarithmic statement.
2 marks
Solve ${log}_{4} x = \frac{3}{2}$ .
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 ${log}_{2} (\frac{1}{16})$ .
2 marks
$\frac{1}{16} = 2^{- 4}$ .
$- 4$
Write $7^{- 1} = \frac{1}{7}$ as a logarithmic statement.
2 marks
The logarithm records the exponent of base 7.
${log}_{7} (\frac{1}{7}) = - 1$
Solve ${log}_{4} x = \frac{3}{2}$ .
2 marks
$x = 4^{\frac{3}{2}} = {(\sqrt{4})}^{3} = 8$ .
$x = 8$
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 ↗

In this lesson

Understand the idea.

A logarithm asks for an exponent

Inverse functions swap input and output

See the method in action.

Read a negative exponent

Translate the logarithm into an exponent.

Express both sides using base 2.

Equate exponents and solve.

Find an inverse with a shifted input

Write the function and its domain and range.

Take a base-3 logarithm and solve for x.

Rename the inverse variables and state its domain.

Try it yourself.

Check your understanding

Keep practising