Solving exponential equations | Year 12

01 / LEARN

Understand the idea.

Isolate, take logarithms, solve

When an exponential term has been isolated, compare bases. Equal powers of the same valid base have equal exponents. Otherwise take logarithms to bring the exponent down as a multiplier, then solve the resulting equation.

The isolated exponential value must be positive. For $a^{m x + n} = c$ , use a base $a > 0$ , $a \neq 1$ , and check $c > 0$ before taking logarithms. With $m \neq 0$ , the remaining equation is linear in $x$ .

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

Look for related powers before using a calculator

Related exponential powers can hide a polynomial. Since $a^{2 x} = {(a^{x})}^{2}$ , substituting $u = a^{x}$ may produce a quadratic. Record $u > 0$ so that impossible polynomial roots are rejected before converting back.

Convert every allowed value of $u$ back to the original variable and check the original equation. Applications may impose further restrictions such as nonnegative time; solving the algebra alone does not establish that a model accepts the solution.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Different bases

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

Worked solution

1
Isolate the exponential term.
$5 \cdot 2^{3 x - 1} = 35 ⟹ 2^{3 x - 1} = 7$
Dividing by 5 leaves a positive right-hand side.
2
Take logarithms and expose the exponent.
$(3 x - 1) ln 2 = ln 7$
The logarithm power law moves the entire exponent outside.
3
Solve the linear equation exactly.
$\begin{aligned} 3 x - 1 & = \frac{ln 7}{ln 2} \\ x & = \frac{1 + \frac{ln 7}{ln 2}}{3} \end{aligned}$
Divide by the nonzero constant ln 2, add 1 and then divide by 3.
4
Round and check the exponential value.
$x \approx 1.269, 2^{3 x - 1} = 2^{\frac{ln 7}{ln 2}} = 7$
The exact exponent reproduces the isolated value, so the original left side is 35.

Answer

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

EXAMPLE 02

Substitute for an exponential

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

Worked solution

1
Substitute for the repeated exponential.
$u = e^{x} > 0, e^{2 x} = u^{2}$
The square of e to the x is e to the 2x.
2
Solve the resulting quadratic.
$u^{2} - 3 u - 4 = (u - 4) (u + 1) = 0 ⟹ u = 4 or u = - 1$
Factoring finds every polynomial candidate.
3
Reject the inadmissible value and convert back.
$u > 0 ⟹ u = 4, e^{x} = 4 ⟹ x = ln 4$
A real exponential cannot equal -1.
4
Check the original equation.
$4^{2} - 3 (4) - 4 = 16 - 12 - 4 = 0$
The remaining value satisfies the original exponential equation.

Answer

$x = ln 4$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = ln 2$

Divide by $3$ to get $e^{2 x} = 4$ . Taking natural logarithms gives $2 x = ln 4 = 2 ln 2$ , so $x = ln 2$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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.
$x = \frac{ln 7}{ln 5}$
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
$e^{- 0.12 t} = \frac{1}{4}$ . Thus $- 0.12 t = - ln 4$ .
$t = \frac{ln 4}{0.12} \approx 11.6$ 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 $e^{- x} = - \frac{1}{7}$ , but real exponentials are positive.
No real solution.

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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.

Isolate, take logarithms, solve

Look for related powers before using a calculator

See the method in action.

Different bases

Isolate the exponential term.

Take logarithms and expose the exponent.

Solve the linear equation exactly.

Round and check the exponential value.

Substitute for an exponential

Substitute for the repeated exponential.

Solve the resulting quadratic.

Reject the inadmissible value and convert back.

Check the original equation.

Try it yourself.

Check your understanding

Keep practising