# Solving exponential and logarithmic equations

> IB Mathematics: Analysis and Approaches SL · Number & Algebra
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u1-solving-exponential-and-logarithmic-equations/

This module covers core methods for solving exponential equations (unknown in the exponent) and logarithmic equations (unknown in a logarithm argument). You will learn to check for invalid solutions and master common exam question formats.

**Prerequisites:** [Laws of logarithms and exponents](https://www.owlsprep.com/study/ib-math-aa-sl-u1-logarithm-laws/); [Properties of exponential functions](https://www.owlsprep.com/study/ib-math-aa-sl-u1-exponential-functions/)

## Learning objectives

- Solve exponential equations using common bases and logarithms
- Solve logarithmic equations using logarithm laws and exponentiation
- Identify and discard extraneous solutions
- Solve quadratic-form exponential and logarithmic equations

## Solving exponential equations by matching bases

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.

**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 $4^{2x - 3} = 32^{1 - x}$ 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 $x$:

   $$9x = 11 \implies x = \frac{11}{9}$$

> **Exam tip**
>
> Always check for matching bases before using logarithms — this saves time and avoids calculation errors.

## Solving exponential equations with different bases

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 $5e^{3x} = 12$, 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:

   $$x = \frac{\ln(2.4)}{3} \approx 0.292$$

> **warning**
>
> Always isolate the exponential term before taking logarithms. Logarithms do not distribute over addition: $\ln(a + b) \neq \ln a + \ln b$.

## Solving logarithmic equations

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

**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 $\log_2(x) + \log_2(x - 2) = 3$ for $x$

1. Use the logarithm product law to combine terms:

   $$\log_2\left(x(x-2)\right) = 3$$
2. Rewrite in exponential form $\log_b A = k \implies A = b^k$:

   $$x(x-2) = 2^3 = 8$$
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: $x=4$

> **Exam tip**
>
> Examiners almost always include an extraneous solution to test whether you check solutions — don't lose easy marks here.

## Quadratic-form exponential and logarithmic equations

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

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

1. Rewrite $2^{2x} = (2^x)^2$, 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 $x=2$

## Common pitfalls

- **Wrong:** Taking the logarithm of a sum before isolating the exponential term: e.g. $\ln(3 + 2^x) = \ln 3 + x\ln 2$
  - Why it fails: Logarithms do not distribute over addition, so this expansion is invalid
  - Correct: Rearrange first to isolate the exponential term on one side of the equation, then take logarithms
- **Wrong:** Forgetting to check for extraneous solutions in logarithmic equations
  - Why it fails: Algebraic rearrangement often produces solutions that make logarithm arguments negative, which is undefined
  - Correct: Always check every solution against the requirement that all logarithm arguments are strictly positive
- **Wrong:** When solving $a^x = b$, writing $x = \frac{\log a}{\log b}$
  - Why it fails: This is the reciprocal of the correct result from misapplying logarithm rules
  - Correct: Take logs of both sides: $x\log a = \log b \implies x = \frac{\log b}{\log a}$
- **Wrong:** Keeping negative solutions for $y = a^x$ when solving quadratic-form equations
  - Why it fails: $a^x$ is always positive for any real $x$ and positive base $a$, so negative values of $y$ cannot produce real solutions
  - Correct: Discard any negative $y$ values before substituting back to solve for $x$

## 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 $y=\log x$, solve quadratic | Discard negative $y=a^x$ solutions |

## 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.

- [Binomial expansion and binomial theorem](https://www.owlsprep.com/study/ib-math-aa-sl-u1-binomial-expansion-and-binomial-theorem/)
- [Functions](https://www.owlsprep.com/study/ib-math-aa-sl-u2-overview/)

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