Exponential equations | Year 10

01 / LEARN

The idea, step by step

Match the base first

For $b > 0$ and $b \neq 1$ , the function $b^{x}$ is one-to-one: if $b^{u} = b^{v}$ , then $u = v$ . Rewriting both sides in the same base turns the equation into a simpler algebra problem.

Divide out an outside coefficient before comparing powers. If $3 (2^{x}) = 24$ , the power equals $8$ , not $24$ .

When there is no convenient shared base

For $b^{x} = c$ with $c > 0$ , $x = log (c) / log (b)$ , using the same log base on the numerator and denominator. A scientific calculator can evaluate this.

Keep the unrounded log expression while checking. An equation such as $2^{x} = - 3$ has no real solution because positive-base exponentials are always positive.

A common trap

From $2^{x} = 16$ , conclude $x = 16 / 2$ .

Instead: The unknown is an exponent. Since $16 = 2^{4}$ , the exponent is $4$ , not $8$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Common base

Solve $4^{x} = 8$ .

Write $4 = 2^{2}$ and $8 = 2^{3}$ .
$2^{2 x} = 2^{3}$ , so $2 x = 3$ and $x = 3 / 2$ .

x = 3 / 2

Example 2

Use a log ratio

Solve $3^{x} = 20$ to 3 decimal places.

$x = log (20) / log (3)$ .
This is approximately $2.726833 \dots$ .
Substitution into $3^{x}$ gives approximately $20$ .

x \approx 2.727

Check your understanding

Solve $5 (2^{x}) = 40$ .

Show answer

$x = 3$

Divide by $5$ to get $2^{x} = 8 = 2^{3}$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $2^{x} = 32$ .
2 marks
Solve $9^{x} = 27$ .
2 marks
Solve $5^{x - 1} = 125$ .
2 marks
Solve ${(1 / 2)}^{x} = 8$ .
2 marks
Solve $2^{x} = 7$ to 3 decimal places.
3 marks
Explain whether $4^{x} + 2 = 1$ has a real solution.
3 marks

Read the worked solutions

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

Solve $2^{x} = 32$ .
2 marks
$32 = 2^{5}$ .
$x = 5$
Solve $9^{x} = 27$ .
2 marks
$3^{2 x} = 3^{3}$ , so $2 x = 3$ .
$x = 3 / 2$
Solve $5^{x - 1} = 125$ .
2 marks
$125 = 5^{3}$ , so $x - 1 = 3$ .
$x = 4$
Solve ${(1 / 2)}^{x} = 8$ .
2 marks
${(1 / 2)}^{x} = 2^{- x}$ and $8 = 2^{3}$ , so $- x = 3$ .
$x = - 3$
Solve $2^{x} = 7$ to 3 decimal places.
3 marks
$x = log 7 / log 2 = 2.807354 \dots$ .
$x \approx 2.807$
Explain whether $4^{x} + 2 = 1$ has a real solution.
3 marks
It would require $4^{x} = - 1$ , but $4^{x}$ is positive for every real $x$ .
No real solution.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Integer and fractional index laws; connecting fractional powers with roots; simplifying surds; operations with surds; rationalising denominators; exact and approximate values; exponential tables and graphs; growth/decay factors; comparing linear/quadratic/exponential patterns; exponential equations; logarithms as inverses; logarithm laws; interpreting logarithmic scales; graph transformations and parameter experiments; choosing and evaluating models.

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