IB Maths AA SL
Paper 1 & 2
17 min read
Solving Equations Analytically
Solving an equation analytically means finding the answer through algebra — no GDC. The strategy depends on whether the unknown appears once (rearrange and apply inverses) or more than once (simplify or use a substitution to get a quadratic).
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What you need to know
- If the unknown appears once: rearrange by applying inverse operations
- If it appears more than once: try simplifying with exponent or log rules first
- If three terms are visible and one is the square of another, try a quadratic substitution
- Watch out for many-to-one operations (squaring, even powers): they can introduce extra false solutions
- Always check your solutions in the original equation
- Never divide by an expression that could be zero — you’ll lose solutions
When the Unknown Appears Once
If x appears in only one place, just peel away the operations by applying their inverses, one at a time:
Common Inverse Pairs
Add
⇄
Subtract
Multiply
⇄
Divide
Reciprocal 1x
⇄
Reciprocal (self-inverse)
Odd power x3
⇄
Cube root 3√x
Even power x2
⇄
±Square root
Exponential ax
⇄
Logarithm logax
ex
⇄
ln x
Many-to-one operations create extra solutions! Even powers and modulus give two answers (±). For example: x + 1 = 3 has one solution (x = 2), but squaring both sides turns it into (x + 1)2 = 9 which has two: x = 2 and x = −4. Always check.
Squaring Adds a Solution
square both sides
creates extra root
(x + 1)2 = 9
→ x = 2 or −4
When the Unknown Appears More Than Once
If x appears in multiple places, the goal is to simplify the expression until only one x remains. Two toolkits help: exponent rules and log rules.
Exponent Rules
af(x) × ag(x) = af(x) + g(x)
af(x) ÷ ag(x) = af(x) − g(x)
(af(x))g(x) = af(x) × g(x)
af(x) = ef(x) ln a
Logarithm Rules
logaf(x) + logag(x) = loga(f(x) × g(x))
logaf(x) − logag(x) = logaf(x)g(x)
n logaf(x) = loga(f(x))n
Common slip: applying a function to each term instead of each side. Starting from ln x + ln(x − 1) = 5, it’s WRONG to write eln x + eln(x−1) = e5. The correct move is eln x + ln(x−1) = e5 (combine the logs first).
The Quadratic Substitution Trick
Some equations look hard but are secretly quadratics in disguise. The clue: three terms involving the same type of function, where one is the square of another.
Example: 2e2x + 3ex − 4 = 0
1
Notice e2x = (ex)2so one term is the square of another
2
Substitute y = exthen e2x becomes y2
3
Now the equation is 2y2 + 3y − 4 = 0solve as a normal quadratic
4
Back-substitute to find xsolve ex = y1 and ex = y2
Spotting the Substitution
Each of these can be turned into a quadratic ay2 + by + c = 0 with the right substitution:
2x4 + 3x2 − 4 = 0
use → y = x2
2e2x + 3ex − 4 = 0
use → y = ex
2 × 25x + 3 × 5x − 4 = 0
use → y = 5x
2x + 3√x − 4 = 0
use → y = √x
Pattern: if you see something like 25x, ask yourself “what’s its square root?” — it’s 5x. So the substitution is y = 5x, because then 25x = (5x)2 = y2.
Watch Out: Dividing by Expressions
If you divide both sides of an equation by an expression, you might lose solutions — because that expression might equal zero!
For example, given (x + 1)(2x − 1) = 3(x + 1):
- Dividing by (x + 1): 2x − 1 = 3 → x = 2
- BUT — x = −1 is also a solution (it makes both sides equal zero!)
The safe approach is to move everything to one side and factorise:
(x + 1)(2x − 1) − 3(x + 1) = 0
(x + 1)(2x − 4) = 0
x = −1 or x = 2 ✓
Worked Examples
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Example 1 — Apply inverses (unknown once)
Solve 5 − 2 log4 x = 0.
Answer:
Step 1: rearrange to isolate log₄ x.
2 log₄ x = 5
log₄ x = 5/2
Step 2: apply the inverse (exponential of base 4).
x = 45/2
= (√4)5 = 25
x = 32
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Example 2 — Square both sides (check for extra solutions!)
Solve x = √(x + 2).
Answer:
Step 1: square both sides (this is many-to-one — beware extra solutions).
x² = x + 2
x² − x − 2 = 0
(x − 2)(x + 1) = 0
x = 2 or x = −1
Step 2: check both solutions in the ORIGINAL equation.
x = 2: LHS = 2, RHS = √4 = 2 ✓
x = −1: LHS = −1, RHS = √1 = 1 ✗ (reject)
x = 2
Squaring created the false solution x = −1. Always check.
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Example 3 — Quadratic substitution (exponential)
Solve e2x − 4ex − 5 = 0.
Answer:
Step 1: notice e2x = (ex)². Substitute y = ex.
y² − 4y − 5 = 0
Step 2: solve the quadratic.
(y + 1)(y − 5) = 0
y = −1 or y = 5
Step 3: back-substitute y = ex.
ex = −1 → no solutions (ex > 0 always)
ex = 5 → x = ln 5
x = ln 5
One of the y values led to no solution — that’s OK, it just means only one x value works.
✎
Example 4 — Combining logarithms
Solve log2(x + 1) + log2(x − 1) = 3.
Answer:
Step 1: combine the logs using the addition rule.
log₂[(x + 1)(x − 1)] = 3
log₂(x² − 1) = 3
Step 2: apply the inverse (exponential of base 2).
x² − 1 = 2³ = 8
x² = 9
x = ±3
Step 3: check the domain — log needs positive arguments.
x = 3: log₂(4) + log₂(2) ✓
x = −3: log₂(−2) — undefined ✗
x = 3
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Example 5 — Don’t divide by a possibly-zero expression
Solve (x − 2)(x + 5) = 4(x − 2).
Answer:
Step 1: move everything to one side. Don’t divide by (x − 2)!
(x − 2)(x + 5) − 4(x − 2) = 0
Step 2: factor out the common bracket (x − 2).
(x − 2)[(x + 5) − 4] = 0
(x − 2)(x + 1) = 0
Step 3: set each factor to zero.
x = 2 or x = −1
If you’d divided by (x − 2), you’d have lost the solution x = 2.
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Tips
- Count how many times x appears. Once → just rearrange. More than once → simplify or substitute.
- Look for the square pattern when there are three terms. If one term is the square of another, you’ve got a hidden quadratic.
- Apply functions to the whole side, not term-by-term. e(ln x + ln y) = e5 is right; eln x + eln y = e5 is wrong.
- Check the domain when logs or square roots are involved. Solutions that make the argument zero or negative aren’t valid.
- Always verify in the original equation, especially after squaring or applying any many-to-one operation.
- Move everything to one side and factorise instead of dividing — you’ll never lose a solution that way.
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Common mistakes
- Applying functions term-by-term. ln x + ln y = 5 → eln x + eln y = e5 is WRONG. Apply the exponential to the whole side.
- Forgetting to check solutions after squaring. Squaring is many-to-one and creates extra (false) solutions.
- Forgetting the domain restrictions for logs and square roots. Valid arguments must be positive.
- Dividing by an expression involving x. If that expression could be zero, you lose solutions. Move things to one side and factorise instead.
- Mistaking e2x for 2ex. They’re different! e2x = (ex)2, not 2 × ex.
- Solving ex = negative number. No real solution exists — ex is always positive. Don’t fake an answer.
- Reversing log and exp: logax = y means x = ay (not ya).
Final word: Look at the structure of the equation. One x? Apply inverses. Multiple x‘s? Simplify. Three terms with a square pattern? Substitute. Then check, check, check — especially when squaring or working with logs.
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