Solving exponential and logarithmic equations

IB Mathematics: Analysis and Approaches SL· Topic 1: Number & Algebra· 45 min read

1. Solving exponential equations by matching bases★★☆☆☆⏱ 15 min

If both sides of an exponential equation can be rewritten to use the same base, we can equate exponents directly, avoiding logarithms entirely. This is the fastest method for these problems.

📘 Definition

Exponential equation

An equation where the unknown variable appears in the exponent of a base.

Example:

$2^{x + 1} = 8^{x}$ is an exponential equation; $x^{2} = 2^{3}$ is not.

📐 Worked Example

Solve $4^{2 x - 3} = 3 2^{1 - x}$ for $x$

1
Rewrite both sides with base 2, since 4 and 32 are powers of 2:
$4 = 2^2, \quad 32 = 2^5$
2
Substitute and apply the exponent rule $(a^{m})^{n} = a^{mn}$ :
$(2^2)^{2x - 3} = (2^5)^{1 - x} \\ 2^{4x - 6} = 2^{5 - 5x}$
3
For $b > 0, b \neq = 1$ , if $b^{m} = b^{n}$ then $m = n$ , so equate exponents:
$4x - 6 = 5 - 5x$
4
Rearrange and solve for $x$ :
$9x = 11 \implies x = \frac{11}{9}$

2. Solving exponential equations with different bases★★★☆☆⏱ 20 min

When bases cannot be rewritten to match, we use the logarithm power rule to bring the exponent down, turning the exponential equation into a linear equation we can solve directly.

📐 Worked Example

Solve $5 e^{3 x} = 12$ , give your answer to 3 significant figures

1
First isolate the exponential term by dividing both sides by 5:
$e^{3x} = \frac{12}{5} = 2.4$
2
Take the natural logarithm of both sides, using $ln (e^{a}) = a$ :
$\ln(e^{3x}) = \ln(2.4) \\ 3x = \ln(2.4)$
3
Solve for $x$ and evaluate:
$x = \frac{\ln(2.4)}{3} \approx 0.292$

3. Solving logarithmic equations★★★☆☆⏱ 20 min

We use the inverse relationship between exponents and logarithms to solve these equations: if $lo g_{b} A = k$ , then $A = b^{k}$ . Because logarithms only have real outputs for positive arguments, we must always check solutions.

📘 Definition

Extraneous solution

A solution that satisfies the rearranged algebraic equation but makes a logarithm argument non-positive in the original equation, so it must be discarded.

📐 Worked Example

Solve $lo g_{2} (x) + lo g_{2} (x - 2) = 3$ for $x$

1
Use the logarithm product law to combine terms:
$\log_2\left(x(x-2)\right) = 3$
2
Rewrite in exponential form $lo g_{b} A = k ⟹ A = b^{k}$ :
$x(x-2) = 2^3 = 8$
3
Rearrange into a standard quadratic equation:
$x^2 - 2x - 8 = 0$
4
Factor and solve:
$(x - 4)(x + 2) = 0 \implies x = 4, x = -2$
5
Final solution: $x = 4$

4. Quadratic-form exponential and logarithmic equations★★★★☆⏱ 20 min

Many equations can be rewritten as quadratics using substitution. Common forms are $a^{2 x} + b a^{x} + c = 0$ (substitute $y = a^{x}$ ) or $(lo g_{b} x)^{2} + c lo g_{b} x + d = 0$ (substitute $y = lo g_{b} x$ ).

📐 Worked Example

Solve $2^{2 x} - 5 (2^{x}) + 4 = 0$ for $x$

1
Rewrite $2^{2 x} = (2^{x})^{2}$ , then substitute $y = 2^{x}$ :
$y^2 - 5y + 4 = 0$
2
Factor the quadratic:
$(y - 1)(y - 4) = 0 \implies y = 1, y = 4$
3
Substitute back $y = 2^{x}$ and solve for $x$ :
$2^x = 1 = 2^0 \implies x = 0 \\ 2^x = 4 = 2^2 \implies x = 2$
4
Check: $2^{x}$ is always positive, so both solutions are valid. Final solutions: $x = 0$ and $x = 2$

5. Common Pitfalls

Wrong move:

Taking the logarithm of a sum before isolating the exponential term: e.g. $ln (3 + 2^{x}) = ln 3 + x ln 2$

Why:

Logarithms do not distribute over addition, so this expansion is invalid

Correct move:

Rearrange first to isolate the exponential term on one side of the equation, then take logarithms

Wrong move:

Forgetting to check for extraneous solutions in logarithmic equations

Why:

Algebraic rearrangement often produces solutions that make logarithm arguments negative, which is undefined

Correct move:

Always check every solution against the requirement that all logarithm arguments are strictly positive

Wrong move:

When solving $a^{x} = b$ , writing $x = \frac{l o g a}{l o g b}$

Why:

This is the reciprocal of the correct result from misapplying logarithm rules

Correct move:

Take logs of both sides: $x lo g a = lo g b ⟹ x = \frac{l o g b}{l o g a}$

Wrong move:

Keeping negative solutions for $y = a^{x}$ when solving quadratic-form equations

Why:

$a^{x}$ is always positive for any real $x$ and positive base $a$ , so negative values of $y$ cannot produce real solutions

Correct move:

Discard any negative $y$ values before substituting back to solve for $x$

6. Quick Reference Cheatsheet

Equation Type	Core Method	Key Check
Exponential, matching bases	Rewrite with same base, equate exponents	None for positive bases
Exponential, different bases	Isolate exponential, take logs, solve for x	Round to required significant figures
Logarithmic equation	Combine logs, rewrite as exponential	All arguments must be positive
Quadratic-in-form	Substitute $y = a^{x}$ or $y = lo g x$ , solve quadratic	Discard negative $y = a^{x}$ solutions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Solve exponential equation
2022 · 2
Solve logarithmic equation
2021 · 1
Quadratic-form exponential equation

Going deeper

syllabusOfficial IB AA SL Topic 1 SyllabusOfficial learning outcome reference

What's Next

Solving exponential and logarithmic equations is a foundational skill for almost all other topics in IB AA SL, from calculus to financial modelling. You will use these skills when solving problems involving exponential growth and decay, compound interest, and when working with derivatives of exponential functions in later calculus units. Mastery of this sub-topic also builds the algebraic manipulation skills you need for more complex equation solving in topics like trigonometry and differential equations. These techniques are regularly tested as part of multi-part questions across both papers, so consistent practice is key.