Solving exponential equations (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“Exponential equations are solved by taking logarithms of both sides and applying the power law to bring the unknown exponent down.”

8 min readUpdated 2 Jun 2026

Quick answer

To solve $a^x=b$ for an exponent, take the logarithm of both sides and use the power law to bring the exponent down: $x\ln a=\ln b$ , so $x=\dfrac{\ln b}{\ln a}$ . The natural log is usually most convenient, especially for equations involving $e$ .

Jump to a section

What this dot point is asking
The core technique
Equations with base $e$
Growth-and-decay applications
Equations that reduce to a quadratic
Choosing which base to use
Common errors
Why it matters
Exam-style questions

What this dot point is asking

When the unknown sits in the exponent, ordinary algebra cannot isolate it. The logarithm is the tool that "undoes" the exponential, letting you bring the exponent down where you can solve for it. This works because of the power law $\log(a^x)=x\log a$ : applying a logarithm converts the exponent into a multiplier, turning an exponential equation into a linear one that ordinary algebra can finish.

The core technique

The strategy

Isolate the exponential term on one side.
Take logs of both sides ( $\ln$ for base $e$ , otherwise any base + change of base).
Use the power law $\log(a^x)=x\log a$ to bring the exponent down.
Solve the resulting linear equation for $x$ .

Solving a base-2 equation

Solve $2^x=20$ .

Step 1: Take the natural log of both sides

The unknown $x$ is trapped in the exponent, so ordinary algebra cannot reach it directly. Taking $\ln$ of both sides creates a valid equation (since $\ln$ is a one-to-one function, both sides remain equal):

\ln(2^x)=\ln 20.

Step 2: Apply the power law to bring the exponent down

The power law of logarithms states $\ln(a^x)=x\ln a$ . Applying it to the left side moves $x$ from the exponent to a multiplier, where it can be isolated:

x\ln 2=\ln 20.

Step 3: Solve for $x$

Divide both sides by $\ln 2$ :

x=\frac{\ln 20}{\ln 2}=\frac{2.9957}{0.6931}\approx 4.322.

Final answer: $x\approx 4.322$ . (Check: $2^{4.322}\approx 20$ , confirming the answer.)

Equations with base $e$

When the base is $e$ , taking $\ln$ is especially clean because $\ln e=1$ .

An equation with e

Solve $150\,e^{-0.3t}=40$ .

Step 1: Isolate the exponential term

Before taking a logarithm, the exponential must be alone on one side. Divide both sides by $150$ :

e^{-0.3t}=\frac{40}{150}\approx 0.2\overline{6}.

This step is essential: taking $\ln$ of both sides before dividing would give $\ln(150\,e^{-0.3t})$ , which requires an extra application of the product law and is a common source of errors.

Step 2: Take the natural log of both sides

Because the base is $e$ , applying $\ln$ is the most direct choice. Since $\ln(e^{u})=u$ for any expression $u$ , the left side simplifies cleanly:

\ln\bigl(e^{-0.3t}\bigr)=\ln(0.2\overline{6}) \implies -0.3t=\ln(0.2667)\approx -1.3218.

Step 3: Solve for $t$

Divide both sides by $-0.3$ :

t=\frac{-1.3218}{-0.3}\approx 4.41.

Final answer: $t\approx 4.41$ .

Growth-and-decay applications

Most exam questions dress this technique as a real model - doubling time, half-life, or reaching a target value. In every case the model has the form $Q=Q_0 e^{kt}$ , where $Q_0$ is the starting amount and $k$ is the continuous growth rate ( $k>0$ ) or decay rate ( $k<0$ ). Reaching a target means setting $Q$ to the target value, isolating the exponential, and taking logs - exactly the technique above. Doubling and half-life are just the special targets $Q=2Q_0$ and $Q=\tfrac12 Q_0$ , where the $Q_0$ cancels and the time depends only on $k$ .

Doubling time

An investment grows as $A=5000\,e^{0.06t}$ (with $t$ in years). How long until it doubles?

Step 1: Set up the equation for doubling

Doubling means the amount reaches twice the starting value, so $A=2\times 5000=10000$ . Substitute into the model:

10000=5000\,e^{0.06t}.

Step 2: Isolate the exponential

Divide both sides by $5000$ . Notice that the starting amount $5000$ cancels, which is why the doubling time depends only on the growth rate $k$ , not on the initial amount:

e^{0.06t}=\frac{10000}{5000}=2.

Step 3: Take the natural log and solve

Apply $\ln$ to both sides. The left side simplifies because $\ln(e^{u})=u$ :

0.06t=\ln 2=0.6931 \implies t=\frac{0.6931}{0.06}\approx 11.55 \text{ years}.

Final answer: The investment doubles after approximately $11.55$ years.

Half-life

A radioactive sample decays as $N=N_0\,e^{-0.045t}$ (days). Find the half-life.

Step 1: Set up the equation for the half-life

The half-life is the time $t$ at which the remaining quantity equals half the starting amount, so set $N=\tfrac{1}{2}N_0$ :

\tfrac{1}{2}N_0=N_0\,e^{-0.045t}.

Step 2: Isolate the exponential

Divide both sides by $N_0$ . The starting amount cancels, confirming that the half-life is a fixed property of the decay rate and does not depend on how much material you begin with:

e^{-0.045t}=0.5.

Step 3: Take the natural log and solve

Apply $\ln$ to both sides. The left side simplifies because $\ln(e^{u})=u$ :

-0.045t=\ln 0.5=-0.6931.

Divide both sides by $-0.045$ . Both sides are negative, so $t$ is positive, as expected:

t=\frac{0.6931}{0.045}\approx 15.4 \text{ days}.

Final answer: The half-life is approximately $15.4$ days.

Equations that reduce to a quadratic

Some exponential equations hide a quadratic. An equation such as $e^{2x}-5e^x+6=0$ becomes a quadratic if you let $u=e^x$ , since $e^{2x}=(e^x)^2=u^2$ . Then $u^2-5u+6=0$ factorises to $(u-2)(u-3)=0$ , giving $u=2$ or $u=3$ . Substituting back, $e^x=2$ gives $x=\ln 2$ and $e^x=3$ gives $x=\ln 3$ . Watch for any value of $u$ that is zero or negative: since $e^x>0$ always, such a value must be rejected because no real $x$ produces it.

Choosing which base to use

You may take logs in any base, but the natural log is almost always the cleanest choice in SACE because the models are written with $e$ and because $\ln e=1$ removes a step. If a question is set in base $10$ or base $2$ , you can still use $\ln$ throughout and the change-of-base ratio falls out automatically: $x=\dfrac{\ln b}{\ln a}$ for $a^x=b$ . The answer is identical whichever base you take; only the intermediate arithmetic differs.

Common errors

Mishandling the log of a difference or dividing too early

You cannot simplify $\ln(A-B)$ , so never split a logarithm of a sum or difference. Also, take logs only after isolating the exponential - applying $\ln$ to $150\,e^{-0.3t}=40$ before dividing by $150$ leads to $\ln(150 e^{-0.3t})$ , which then needs the product law and is a common source of errors. Isolate first, then take logs.

Keep extra decimal places until the final line

Logarithms and divisions amplify rounding. Carry at least four decimal places through intermediate steps and round only the final answer, or your doubling time/half-life can be off in the last figure. State the degree of accuracy the question asks for.

Why it matters

Solving exponential equations is the payoff of Topic 4 and a guaranteed exam skill, especially in growth-and-decay contexts. It draws together the log laws, change of base, and the exponential models you differentiated earlier, and it recurs whenever a continuous random variable or model must be inverted for a target value.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style3 marks

Calculator-assumed. A colony of bacteria is modelled by

N = 800\,e^{0.25t}

, where

t

is in hours. Find, to the nearest tenth of an hour, the time for the colony to reach

5000

Show worked answer →

Set $N = 5000$ and isolate the exponential:

5000 = 800\,e^{0.25t} \quad\Rightarrow\quad e^{0.25t} = \frac{5000}{800} = 6.25.

Take natural logs:

0.25t = \ln 6.25 \approx 1.8326 \quad\Rightarrow\quad t = \frac{1.8326}{0.25} \approx 7.3 \text{ hours}.

Marks: one for isolating $e^{0.25t} = 6.25$ , one for taking logs, one for $t \approx 7.3$ . Dividing by $800$ before taking logs is essential.

SACE 2022-style3 marks

Calculator-assumed. Solve

3^{2x - 1} = 50

, giving

x

to three decimal places.

Show worked answer →

Take natural logs of both sides and apply the power law:

(2x - 1)\ln 3 = \ln 50 \quad\Rightarrow\quad 2x - 1 = \frac{\ln 50}{\ln 3} = \frac{3.9120}{1.0986} \approx 3.5609.

Then $2x = 4.5609$ , so $x \approx 2.280$ .

Marks: one for taking logs with the power law, one for $\dfrac{\ln 50}{\ln 3}$ , one for solving to $x \approx 2.280$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

The core technique

Step 1: Take the natural log of both sides

Step 2: Apply the power law to bring the exponent down

Step 3: Solve for xxx

Equations with base eee

Step 1: Isolate the exponential term

Step 2: Take the natural log of both sides

Step 3: Solve for ttt

Growth-and-decay applications

Step 1: Set up the equation for doubling

Step 2: Isolate the exponential

Step 3: Take the natural log and solve

Step 1: Set up the equation for the half-life

Step 2: Isolate the exponential

Step 3: Take the natural log and solve

Equations that reduce to a quadratic

Choosing which base to use

Common errors

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Step 3: Solve for $x$

Equations with base $e$

Step 3: Solve for $t$