Solving exponential equations with base-10 logs | Year 11

01 / LEARN

Understand the idea.

Isolate the exponential term

First isolate the expression containing the unknown exponent. If matching bases is inconvenient, take the same logarithm on both positive sides. The power law then brings the exponent outside, leaving an equation that can usually be solved by ordinary algebra.

Check that the base is positive and different from 1 before dividing by its logarithm. Retain the full calculator expression through the final division; rounding the individual logarithms can noticeably change the solution.

b^{m x + c} = k ⟹ x = \frac{log k / log b - c}{m}

Interpret a time solution

A model equation may produce a real-valued crossing time. If the question instead asks for the first completed period, test the nearby whole-number periods in the original condition. Rounding a crossing time normally is not sufficient for a threshold question.

For a decay factor between zero and one, its logarithm is negative. Dividing an inequality by that logarithm reverses its direction. Substitute the chosen time into the original model to confirm which side of the threshold it reaches.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A linear expression in an exponent

Solve $3^{2 x - 1} = 20$ , to four decimal places.

Worked solution

1
Take common logarithms of the positive sides.
$log (3^{2 x - 1}) = log 20$
Both sides are positive, so the same one-to-one logarithm preserves equality.
2
Bring the exponent outside and divide.
$(2 x - 1) log 3 = log 20 ⟹ 2 x - 1 = \frac{log 20}{log 3}$
The power law applies, and $log 3 \neq 0$ .
3
Isolate the variable.
$x = \frac{1}{2} (1 + \frac{log 20}{log 3})$
Adding 1 and then dividing by 2 undoes the linear expression in the exponent.
4
Evaluate to the requested precision.
$x \approx 1.8634$
Evaluate the whole expression before rounding to four decimal places.

Answer

x approximately 1.8634

EXAMPLE 02

Isolate a decay term

Solve $5 - 2 {(0.7)}^{t} = 4$ for t.

Worked solution

1
Isolate the exponential factor.
$5 - 2 {(0.7)}^{t} = 4 ⟹ 2 {(0.7)}^{t} = 1 ⟹ {(0.7)}^{t} = 0.5$
Subtracting and dividing must be completed before using the logarithm power law.
2
Take common logarithms.
$t log 0.7 = log 0.5$
Both exponential values are positive, so their real logarithms exist.
3
Divide and evaluate.
$t = \frac{log 0.5}{log 0.7} \approx 1.9434$
Both logarithms are negative, giving the expected positive time.
4
Check in the original expression.
$5 - 2 (0.5) = 4$
The solved exponential factor returns exactly the required output.

Answer

t approximately 1.9434

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $4^{x} = 9$ using base-ten logs.

Show answer & reasoning

$x = log 9 / log 4 \approx 1.5850$

Taking common logarithms gives $x log 4 = log 9$ . Since $log 4 \neq 0$ , $x = \frac{log 9}{log 4} \approx 1.5850$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Solve $8^{x} = 12$ to four decimal places.
2 marks
Solve $2^{x + 3} = 7$ to four decimal places.
2 marks
Solve $4 {(1.2)}^{t} = 10$ to three decimal places.
2 marks
A 600-unit quantity grows by 8% at the end of each year. In which completed year does it first exceed 1000?
3 marks
Solve ${(0.8)}^{t} < 0.3$ for real $t$ .
3 marks
Can $2^{x} = - 3$ have a real solution? Explain.
2 marks

Read the worked solutions

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

Solve $8^{x} = 12$ to four decimal places.
2 marks
$x = \frac{log 12}{log 8}$ .
$x \approx 1.1950$
Solve $2^{x + 3} = 7$ to four decimal places.
2 marks
$x + 3 = \frac{log 7}{log 2}$ , then subtract 3.
$x \approx - 0.1926$
Solve $4 {(1.2)}^{t} = 10$ to three decimal places.
2 marks
Isolate ${(1.2)}^{t} = 2.5$ ; use $t = \frac{log 2.5}{log 1.2}$ .
$t \approx 5.026$
A 600-unit quantity grows by 8% at the end of each year. In which completed year does it first exceed 1000?
3 marks
$t > \frac{log (\frac{1000}{600})}{log 1.08} \approx 6.637$ .
Year 7 is the first integer year; substitution verifies the threshold.
7
Solve ${(0.8)}^{t} < 0.3$ for real $t$ .
3 marks
Take logs and divide by $log 0.8 < 0$ , reversing the inequality.
$t > \frac{log 0.3}{log 0.8} \approx 5.396$
Can $2^{x} = - 3$ have a real solution? Explain.
2 marks
$2^{x}$ is positive for every real $x$ , so it cannot equal a negative number.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

In this lesson

Understand the idea.

Isolate the exponential term

Interpret a time solution

See the method in action.

A linear expression in an exponent

Take common logarithms of the positive sides.

Bring the exponent outside and divide.

Isolate the variable.

Evaluate to the requested precision.

Isolate a decay term

Isolate the exponential factor.

Take common logarithms.

Divide and evaluate.

Check in the original expression.

Try it yourself.

Check your understanding

Keep practising