Solving Equations Analytically (DP IB Analysis & Approaches (AA)): Revision Note

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on 18 July 2025

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Solving Equations Analytically

How can I solve equations analytically where the unknown appears only once?

These equations can be solved by rearranging
For one-to-one functions you can just apply the inverse
- Addition and subtraction are inverses
  - $y = x + 3 ⟺ x = y - 3$
- Multiplication and division are inverses
  - $y = 3 x ⟺ x = \frac{y}{3}$
- Taking the reciprocal is a self-inverse
  - $y = \frac{1}{x} ⟺ x = \frac{1}{y}$
- Odd powers and roots are inverses
  - $y = x^{3} ⟺ x = \sqrt[3]{y} = y^{\frac{1}{3}}$
- Exponentials and logarithms are inverses
  - $y = 3^{x} ⟺ x = \log_{3} y$
  - $y = e^{x} \Leftrightarrow x = \ln y$
For many-to-one functions you will need to use your knowledge of the functions to find the other solutions
- Even powers lead to positive and negative solutions
  - $y = x^{4} \Leftrightarrow x = \pm \sqrt[4]{y}$
- Modulus functions lead to positive and negative solutions
  - $y = |x| \Leftrightarrow x = \pm y$
- Trigonometric functions lead to infinite solutions using their symmetries
  - $y = \sin x \Leftrightarrow x = 2 k π + \sin^{- 1} y or x = (1 + 2 k) π - \sin^{- 1} y$
  - $y = \cos x \Leftrightarrow x = 2 k π \pm \cos^{- 1} y$
  - $y = \tan x \Leftrightarrow x = k π + \tan^{- 1} y$
Take care when you apply many-to-one functions to both sides of an equation as this can create additional solutions which are incorrect
- For example: squaring both sides
  - $x + 1 = 3$ has one solution $x = 2$
  - ${(x + 1)}^{2} = 3^{2}$ has two solutions $x = 2$ and $x = - 4$
- Always check your solutions by substituting back into the original equation

How can I solve equations analytically where the unknown appears more than once?

Sometimes it is possible to simplify expressions to make the unknown appear only once
Collect all terms involving x on one side and try to simplify into one term
- For exponents use
  - $a^{f (x)} \times a^{g (x)} = a^{f (x) + g (x)}$
  - $\frac{a^{f (x)}}{a^{g (x)}} = a^{f (x) - g (x)}$
  - ${(a^{f (x)})}^{g (x)} = a^{f (x) \times g (x)}$
  - $a^{f (x)} = e^{f (x) \ln a}$
- For logarithms use
  - $\log_{a} f (x) + \log_{a} g (x) = \log_{a} (f (x) \times g (x))$
  - $\log_{a} f (x) - \log_{a} g (x) = \log_{a} (\frac{f (x)}{g (x)})$
  - $n \log_{a} f (x) = \log_{a} {(f (x))}^{n}$

How can I solve equations analytically when the equation can't be simplified?

Sometimes it is not possible to simplify equations
Most of these equations cannot be solved analytically
A special case that can be solved is where the equation can be transformed into a quadratic using a substitution
- These will have three terms and involve the same type of function
Identify the suitable substitution by considering which function is a square of another
- For example: the following can be transformed into $2 y^{2} + 3 y - 4 = 0$
  - $2 x^{4} + 3 x^{2} - 4 = 0$ using $y = x^{2}$
  - $2 x + 3 \sqrt{x} - 4 = 0$ using $y = \sqrt{x}$
  - $\frac{2}{x^{6}} + \frac{3}{x^{3}} - 4 = 0$ using $y = \frac{1}{x^{3}}$
  - $2 e^{2 x} + 3 e^{x} - 4 = 0$ using $y = e^{x}$
  - $2 \times 25^{x} + 3 \times 5^{x} - 4 = 0$ using $y = 5^{x}$
  - $2^{2 x + 1} + 3 \times 2^{x} - 4 = 0$ using $y = 2^{x}$
  - $2 {(x^{3} - 1)}^{2} + 3 (x^{3} - 1) - 4 = 0$ using $y = x^{3} - 1$
To solve:
- Make the substitution $y = f (x)$
- Solve the quadratic equation $a y^{2} + b y + c = 0$ to get y₁ & y₂
- Solve $f (x) = y_{1}$ and $f (x) = y_{2}$
  - Note that some equations might have zero or several solutions

Can I divide both sides of an equation by an expression?

When dividing by an expression you must consider whether the expression could be zero
Dividing by an expression that could be zero could result in you losing solutions to the original equation
- For example: $(x + 1) (2 x - 1) = 3 (x + 1)$
  - If you divide both sides by $(x + 1)$ you get $2 x - 1 = 3$ which gives $x = 2$
  - However $x = - 1$ is also a solution to the original equation
To ensure you do not lose solutions you can:
- Split the equation into two equations
  - One where the dividing expression equals zero: $x + 1 = 0$
  - One where the equation has been divided by the expression: $2 x - 1 = 3$
- Make the equation equal zero and factorise
  - $(x + 1) (2 x - 1) - 3 (x + 1) = 0$
  - $(x + 1) (2 x - 1 - 3) = 0$ which gives $(x + 1) (2 x - 4) = 0$
  - Set each factor equal to zero and solve: $x + 1 = 0$ and $2 x - 4 = 0$

Examiner Tips and Tricks

A common mistake that students make in exams is applying functions to each term rather than to each side

For example: Starting with the equation $\ln x + \ln (x - 1) = 5$ it would be incorrect to write $e^{\ln x} + e^{\ln (x - 1)} = e^{5}$ or $x + (x - 1) = e^{5}$
Instead it would be correct to write $e^{\ln x + \ln (x - 1)} = e^{5}$ and then simplify from there

Worked Example

Find the exact solutions for the following equations:

a) $5 - 2 \log_{4} x = 0$ .

2-4-3-ib-aa-sl-solve-analytically-a-we-solution

b) $x = \sqrt{x + 2}$ .

2-4-3-ib-aa-sl-solve-analytically-b-we-solution

c) $e^{2 x} - 4 e^{x} - 5 = 0$ .

2-4-3-ib-aa-sl-solve-analytically-c-we-solution

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Solving Equations Analytically (DP IB Analysis & Approaches (AA)): Revision Note

Solving Equations Analytically

How can I solve equations analytically where the unknown appears only once?

How can I solve equations analytically where the unknown appears more than once?

How can I solve equations analytically when the equation can't be simplified?

Can I divide both sides of an equation by an expression?

You've read 0 of your 5 free revision notes this week

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Join the 100,000+ Students that ❤️ Save My Exams

1. Number & Algebra

Number Toolkit

Standard Form

Laws of Indices

Exponentials & Logs

Introduction to Logarithms

Laws of Logarithms

Solving Exponential Equations

Sequences & Series

Language of Sequences & Series

Sigma Notation

Arithmetic Sequences & Series

Geometric Sequences & Series

Applications of Sequences & Series

Compound Interest & Depreciation

Proof & Reasoning

Proof

Binomial Theorem

Binomial Coefficients & Pascal's Triangle

Binomial Theorem

2. Functions

Linear Functions & Graphs

Equations of a Straight Line

Parallel & Perpendicular Lines

Quadratic Functions & Graphs

Quadratic Functions

Factorising Quadratics

Completing the Square

Solving Quadratic Equations

Quadratic Inequalities

Discriminants

Functions Toolkit

Language of Functions

Composite Functions

Inverse Functions

Graphing Functions & Their Key Features

Intersecting Graphs

Further Functions & Graphs

Reciprocal & Rational Functions

Exponential & Logarithmic Functions

Solving Equations Analytically

Solving Equations Graphically

Modelling with Functions

Transformations of Graphs

Translations of Graphs

Reflections of Graphs

Stretches of Graphs

Composite Transformations of Graphs

3. Geometry & Trigonometry

Geometry Toolkit

Coordinate Geometry

Arcs & Sectors Using Degrees

Radian Measure

Arcs & Sectors Using Radians

Geometry of 3D Shapes

3D Coordinate Geometry

Volume & Surface Area

Trigonometry

Pythagoras & Right-Angled Trigonometry

Sine Rule, Cosine Rule & Area of a Triangle

Angles of Elevation & Depression

Bearings & Constructions

The Unit Circle & Exact Values

The Unit Circle

Exact Values

Trigonometric Functions & Graphs

Graphs of Trigonometric Functions

Solving Equations Using Trigonometric Graphs

Transformations of Trigonometric Functions

Modelling with Trigonometric Functions

Trigonometric Equations & Identities