Mathematical Methods · Units 3 & 4

Solving Exponential and Logarithmic Equations

Learn to solve exponential and logarithmic equations the easy way, with plain English intuition, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Stuck. That is how an unknown feels when it is trapped up in a power, like the $x$ in $2^x = 7$ . You cannot just divide it out or move it across. So mathematicians built a special key that pulls a power back down to ground level where you can deal with it. That key is the logarithm, and once you know how to turn it, every exponential equation springs open.

The unknown is hiding in the power

A normal equation like $2x = 7$ is easy because $x$ sits out in the open. You divide by $2$ and you are done. But $2^x = 7$ is different. Here the $x$ is the power, and ordinary algebra has no move to bring it down. You need an operation that undoes “raising to a power”.

That operation is the logarithm. The rule that does the heavy lifting is

\log_e(a^x) = x\log_e(a).

Read it right to left and it is magic. Taking $\log_e$ of both sides of $2^x = 7$ slides the $x$ from the power down to the front, where it becomes a plain multiplier you can divide away.

The standard recipe for $a^x = b$

Once you see the pattern, solving $a^x = b$ is a short routine.

Get the power on its own first. If there is a number multiplying it, divide that away before anything else.
Take $\log_e$ of both sides.
Use $\log_e(a^x) = x\log_e(a)$ to bring the power down.
Divide by $\log_e(a)$ to leave $x$ by itself.

The most common trap is doing things in the wrong order. If the equation is $5 \times 2^x = 40$ , you must divide by $5$ to reach $2^x = 8$ before you ever touch a logarithm. Take logs too early and the working becomes a tangle.

When a quadratic is hiding inside

Some equations look scary, like $e^{2x} - 5e^x + 6 = 0$ , but they are quadratics in disguise. The clue is that $e^{2x}$ is just $\left(e^x\right)^2$ . Let $a = e^x$ and the whole thing becomes $a^2 - 5a + 6 = 0$ , which factorises in the usual way.

After you solve for $a$ , swap back to $e^x$ and finish. Here is the step that separates the careful student from the rest. An exponential like $e^x$ is always positive, so if your quadratic hands you a value such as $e^x = -1$ , that branch has no solution and you throw it away. Examiners report this rejection step as one of the most frequently missed marks.

The curve y = e^x stays strictly above the x-axis, so e^x can never equal a negative number like -1, which is why that branch is rejected.

Learn it, then fade the help

1 · I do — fully worked

A hidden quadratic with a rejection Solve $e^{2x} - 5e^x + 6 = 0$ . Let $a = e^x$ , so $a^2 - 5a + 6 = 0$ , giving $(a - 2)(a - 3) = 0$ , so $a = 2$ or $a = 3$ . Both are positive, so $e^x = 2$ or $e^x = 3$ , giving $x = \log_e(2)$ or $x = \log_e(3)$ .

2 · We do — your turn, guided

Same steps, your turn. Solve $e^{2x} - 3e^x - 4 = 0$ . Let $a = e^x$ to get $a^2 - 3a - 4 = 0$ , which factorises to $(a - 4)(a + 1) = 0$ , so $a = \;?$ — then reject any value that breaks $e^x > 0$ .

Check your step

$a = 4$ or $a = -1$ . Since $e^x > 0$ , reject $a = -1$ . So $e^x = 4$ , giving $x = \log_e(4)$ .

3 · You do — solo

No scaffolding now: solve $e^{2x} - 12 = 4e^x$ for $x \in R$ .

Reveal solution

Let $a = e^x$ : $a^2 - 4a - 12 = 0$ , so $(a - 6)(a + 2) = 0$ , giving $a = 6$ or $a = -2$ . Reject $a = -2$ (since $e^x > 0$ ). So $e^x = 6$ , giving $x = \log_e(6)$ .

Solving logarithmic equations

A logarithmic equation has the unknown trapped inside a log instead. The plan is to collapse everything into a single log, then strip the logs away.

If you can write the equation as $\log_e(\text{something}) = \log_e(\text{something else})$ , then the two insides must be equal, and the logs vanish. The log laws you lean on are

\log_e(a) + \log_e(b) = \log_e(ab), \qquad \log_e(a) - \log_e(b) = \log_e\!\left(\tfrac{a}{b}\right).

There is one rule you can never skip. A logarithm only accepts a positive input, so $\log_e(x)$ needs $x > 0$ and $\log_e(x - 3)$ needs $x > 3$ . Solve the equation, then check every answer against this implied domain and discard any that break it.

The check that wins marks

Both kinds of equation share the same final move, and it is the one students skip under pressure. After exponentials, reject any branch that asks an exponential to be zero or negative, because $e^x > 0$ for every $x$ . After logarithms, substitute each answer back and reject any that makes a log argument zero or negative. A solution that survives this check is a real solution. One that does not is an impostor that the algebra invented along the way.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

The unknown is stuck in a power. What is your first move?

Write the exact solution of

a^x = b

5 \times 2^x = 40

, what must you do before taking logs?

How do you solve

e^{2x} - 5e^x + 6 = 0

, and what extra check is needed?

Why must you check the domain after solving a log equation?

Recall · Index and Logarithm Laws

Rewrite

\log_b(y) = x

in exponential form.

Recall · Solving Polynomial Equations

State the null factor law used to finish these quadratics.

See the full recipes in the Worked Examples tab, then test yourself in Try It.

Worked Example 1A logarithmic equation hiding a cubic, from a real exam

Solve $2\log_3(x - 4) + \log_3(x) = 2$ for $x$ .

1
Combine the logs using the power and product laws, noting the domain $x > 4$ .
$\log_3\!\left((x-4)^2 x\right) = 2 \;\Rightarrow\; x(x-4)^2 = 9$
2
Expand and factorise the cubic.
$x^3 - 8x^2 + 16x - 9 = (x-1)(x^2 - 7x + 9) = 0$
3
Solve, then reject the roots outside the domain $x > 4$ (here $x = 1$ and $x = \tfrac{7 - \sqrt{13}}{2}$ ).
$x = \frac{7 + \sqrt{13}}{2}$

Answer

x = \dfrac{7 + \sqrt{13}}{2}

VCAA 2024 Mathematical Methods Exam 1, Q6

Worked Example 2A quadratic in $e^x$ with a root to reject, from a real exam

Solve $e^{2x} - 12 = 4e^x$ for $x \in R$ .

1
Let $a = e^x$ and rearrange into a quadratic equal to zero.
$a^2 - 4a - 12 = 0$
2
Factorise and solve for $a$ .
$(a-6)(a+2) = 0 \implies a = 6 \text{ or } a = -2$
3
Since $e^x > 0$ , reject $a = -2$ and take logs of $e^x = 6$ .
$x = \log_e(6)$

Answer

x = \log_e(6)

VCAA 2023 Mathematical Methods Exam 1, Q2

Worked Example 3Bring the power down

Solve $2^x = 7$ for $x$ , giving an exact answer.

1
The unknown is stuck up in the power, so take $\log_e$ of both sides.
$\log_e(2^x) = \log_e(7)$
2
The log law $\log_e(a^x) = x\log_e(a)$ slides the power down to the front.
$x\log_e(2) = \log_e(7)$
3
Divide both sides by $\log_e(2)$ to free $x$ .
$x = \frac{\log_e(7)}{\log_e(2)}$

Answer

x = \dfrac{\log_e(7)}{\log_e(2)}

Worked Example 4A hidden quadratic

Solve $e^{2x} - 5e^{x} + 6 = 0$ for $x$ .

1
Let $a = e^x$ . Then $e^{2x} = (e^x)^2 = a^2$ , so it is just a quadratic in $a$ .
$a^2 - 5a + 6 = 0$
2
Factorise.
$(a-2)(a-3) = 0 \quad\Rightarrow\quad a = 2 \text{ or } a = 3$
3
Swap $a$ back to $e^x$ . Both values are positive, so both are valid since $e^x > 0$ .
$e^x = 2 \quad\text{or}\quad e^x = 3$
4
Take $\log_e$ of each side to release $x$ .
$x = \log_e(2) \quad\text{or}\quad x = \log_e(3)$

Answer

x = \log_e(2) \text{ or } x = \log_e(3)

Worked Example 5A log equation with a trap

Solve $\log_e(x) + \log_e(x-3) = \log_e(10)$ for $x$ .

1
Combine the left side with the law $\log_e(a) + \log_e(b) = \log_e(ab)$ .
$\log_e\big(x(x-3)\big) = \log_e(10)$
2
Equal logs mean equal insides. Drop the logs.
$x(x-3) = 10$
3
Expand and rearrange into a standard quadratic.
$x^2 - 3x - 10 = 0$
4
Factorise and solve.
$(x-5)(x+2) = 0 \quad\Rightarrow\quad x = 5 \text{ or } x = -2$
5
Check the domain. The original equation needs $x > 0$ and $x - 3 > 0$ , so $x > 3$ . Reject $x = -2$ .
$x = 5$

Answer

x = 5

Practice test

Try it yourself

10 questions, 15 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.The points $P$ and $Q$ lie on the line $y = x$ and on the graph of $g(x) = \tfrac12\big(e^{2-x} + e^{-(2-x)}\big)$ . Find the coordinates of $P$ and $Q$ , correct to two decimal places.

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2023 Mathematical Methods Exam 2, Section B Q5ci

Q2.Solve $e^{2x} - 8e^{x} + 7 = 0$ for $x$ .

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Mathematical Methods Exam 1, Q5a

Q3.Find the time $t \in (0, 1)$ when the temperatures predicted by the models $f$ and $g$ are equal, where $f(t) = 12 + 30t$ and $g(t) = 22 - 10e^{-6t}$ on the relevant interval. Give your answer correct to two decimal places.

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2024 Mathematical Methods Exam 2, Section B Q2d

Q4.Let $f:R\to R,\ f(x)=\dfrac{x}{2}+7$ and $g:R\to R,\ g(x)=Ae^{kx}$ where $A,k\in R$ . The graphs intersect at $(-12,1)$ and $(2,8)$ . Write down two simultaneous equations in terms of $A$ and $k$ , and solve them using algebra to show that $A=2^{\frac{18}{7}}$ and $k=\dfrac{3}{14}\log_e(2)$ .

3marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Mathematical Methods Exam 2, Section B Q2a

Q5.The exact solution of $3^x = 20$ is:

1mark

Need a hint?

Take

\log_e

of both sides, bring the power down, then divide by

\log_e(3)

rather than subtracting.

Q6.The equation $e^{2x} - 3e^{x} - 4 = 0$ has solution set:

1mark

Need a hint?

Let

a = e^x

to get a quadratic, then remember that

e^x

can never be negative.

Q7.The solution of $\log_e(x) + \log_e(x+2) = \log_e(15)$ is:

1mark

Need a hint?

Combine the two logs into one, solve the quadratic, then check each root against

x > 0

Q8.The exact solution of $5 \times 2^{x} = 40$ is:

1mark

Need a hint?

Divide by

5

to isolate the power before doing anything else, then see if the right side is a neat power of

2

Q9.Over the real numbers, the equation $\log_e(x - 4) = 1$ has solution:

1mark

Need a hint?

Rewrite the log equation in exponential form:

\log_e(\text{something}) = 1

means the something equals

e

Q10.Solve $\log_e(2x + 3) + \log_e(x) = \log_e(9)$ for $x$ . Show every step and justify your final answer.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

How do you solve an equation when x is in the exponent?

Get the power by itself, then take the natural log of both sides. The log law lets you bring the exponent down to the front as a multiplier, and from there you divide to leave x on its own.

Why do you reject some solutions to a logarithmic equation?

A logarithm only accepts a positive input, so the original equation has an implied domain. After solving, you substitute each answer back and discard any that would make a log argument zero or negative, since those are not real solutions.

How do you solve an exponential equation that looks like a quadratic?

Substitute a single letter for the exponential term, such as a for e to the x, which turns it into an ordinary quadratic you can factorise. After solving, swap back and remember that an exponential is always positive, so reject any negative value.

Solving Exponential and Logarithmic Equations

Learn

The unknown is hiding in the power

The standard recipe for ax=ba^x = bax=b

When a quadratic is hiding inside

Solving logarithmic equations

The check that wins marks

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions

The standard recipe for $a^x = b$