Solving exponential equations | Year 12

01 / LEARN

The idea, step by step

Isolate, take logarithms, solve

If two positive expressions have the same base, equate their exponents. Otherwise first isolate the exponential term. Taking logarithms turns its exponent into a multiplier, after which an ordinary equation can be solved. An exponential equation such as $a^{m x + n} = c$ has a real solution only if its isolated right side $c$ is positive.

a^{m x + n} = c ⟹ x = \frac{ln c / ln a - n}{m} (c > 0, m \neq 0)

Look for related powers before using a calculator

When both $a^{x}$ and $a^{2 x}$ appear, set $u = a^{x}$ , remembering $u > 0$ . Solve the resulting polynomial in $u$ , reject nonpositive values, and convert back to $x$ . In applications, also check a time restriction such as $t \geq 0$ ; an algebraic answer can lie outside the model.

A common trap

Taking $ln (3 + 2^{x})$ as $ln 3 + x ln 2$ .

Instead: First isolate the exponential term. Logarithm laws separate products and quotients, not sums.

02 / WORKED EXAMPLES

See the thinking

Example 1

Different bases

Solve $5 \cdot 2^{3 x - 1} = 35$ .

Divide by $5$ : $2^{3 x - 1} = 7$ .
Take natural logarithms: $(3 x - 1) ln 2 = ln 7$ .
Solve the linear equation in $x$ .

x = \frac{1 + ln 7 / ln 2}{3} \approx 1.269

Example 2

Substitute for an exponential

Solve $e^{2 x} - 3 e^{x} - 4 = 0$ .

Put $u = e^{x} > 0$ . Then $u^{2} - 3 u - 4 = (u - 4) (u + 1) = 0$ .
Reject $u = - 1$ because $e^{x} > 0$ .
The remaining equation is $e^{x} = 4$ .

x = ln 4

Check your understanding

Solve $3 e^{2 x} = 12$ exactly.

Show answer

$x = ln 2$

$e^{2 x} = 4$ , so $2 x = ln 4 = 2 ln 2$ .

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $3^{2 x - 1} = 27$ .
2 marks
Solve $4 \cdot 5^{x} = 28$ exactly.
2 marks
Solve $e^{2 x} - 5 e^{x} + 6 = 0$ .
3 marks
A quantity is $Q (t) = 80 e^{- 0.12 t}$ , with $t$ in hours. When is $Q = 20$ ?
3 marks
Solve $2^{x} = 3^{x - 1}$ .
3 marks
Explain whether $7 e^{- x} + 2 = 1$ has a real solution.
2 marks

Read the worked solutions

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

Solve $3^{2 x - 1} = 27$ .
2 marks
$27 = 3^{3}$ , so $2 x - 1 = 3$ .
$x = 2$
Solve $4 \cdot 5^{x} = 28$ exactly.
2 marks
$5^{x} = 7$ , then take logarithms.
$x = ln 7 / ln 5$
Solve $e^{2 x} - 5 e^{x} + 6 = 0$ .
3 marks
Let $u = e^{x}$ . $(u - 2) (u - 3) = 0$ and both roots are positive.
$x = ln 2$ or $x = ln 3$
A quantity is $Q (t) = 80 e^{- 0.12 t}$ , with $t$ in hours. When is $Q = 20$ ?
3 marks
$e^{- 0.12 t} = 1 / 4$ . Thus $- 0.12 t = - ln 4$ .
$t = ln 4 / 0.12 \approx 11.6$ hours
Solve $2^{x} = 3^{x - 1}$ .
3 marks
$x ln 2 = (x - 1) ln 3$ . Collect the terms in $x$ .
$x = \frac{ln 3}{ln 3 - ln 2} \approx 2.710$
Explain whether $7 e^{- x} + 2 = 1$ has a real solution.
2 marks
It would require $e^{- x} = - 1 / 7$ , but real exponentials are positive.
No real solution.

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Natural logarithms

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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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