IB Maths AA HL
Topic 2 — Functions
Paper 1 & 2
~9 min read
Solving Equations Analytically
Solving an equation analytically means using algebra alone — no GDC, no graph. The recipe is short: isolate the unknown by undoing each operation, one at a time, using its inverse. The catches are when the unknown appears more than once (combine first), when many-to-one functions create extra “fake” solutions (always check), and when the equation has the shape of a hidden quadratic (substitute and reveal). This note covers all three.
📘 What you need to know
- Unknown appears once: rearrange by applying inverses one at a time. Add ↔ Subtract, Multiply ↔ Divide, Power ↔ Root, Exponential ↔ Logarithm.
- Many-to-one inverses produce extra solutions: even powers give ±, modulus gives ±, trig gives infinitely many. Always check by substitution.
- Unknown appears more than once: try to combine using exponent rules or log rules until the unknown shows up only once. Then rearrange.
- Hidden quadratic: if the equation has three terms where one is the square of another (e.g. e2x + ex + 1), substitute u = the simpler term and solve as a quadratic.
- Exam exact values: questions asking for “exact” answers want you to leave logs, surds, or fractions in your answer — don’t decimalise.
- Apply functions to entire sides, not term-by-term. eln x + ln(x−1) ≠ x + (x − 1).
- Don’t divide by an expression that could be zero — you’ll lose solutions. Subtract instead and factor.
The inverse toolkit
Every operation has an inverse. To isolate x, apply the inverse of whatever’s been done to it — but apply it to the whole side, not just one term:
add ↔ subtract
y = x + 4 ⟺ x = y − 4
subtract to undo addition
× ↔ ÷
y = 5x ⟺ x = y/5
divide to undo multiplication
power ↔ root
y = x3 ⟺ x = ∛y
odd powers: one inverse, no ±
exp ↔ log
y = ex ⟺ x = ln y
take ln to bring down the exponent
log ↔ exp
y = loga x ⟺ x = ay
exponentiate to undo a log
reciprocal (self)
y = 1/x ⟺ x = 1/y
flip both sides
One-to-one functions are safe: their inverse gives exactly one value of x. Even powers, modulus, and trig functions are many-to-one — their inverses spawn extra cases (±, multiple solutions) which you must track explicitly.
The many-to-one trap — extra solutions
Some operations are many-to-one, meaning two different inputs give the same output. When you “undo” them, you create two cases:
Even powers
x2 = 25 ⟹ x = ±5
always include both signs
Modulus
|x| = 7 ⟹ x = ±7
two values, both distance 7 from 0
Squaring both sides is a powerful move but it can also create false solutions. After squaring, always check by plugging your solutions back into the original equation. If a value doesn’t satisfy the original, throw it out.
When the unknown appears more than once
Combine into one term first, then rearrange. The two main tricks:
Combining exponents
Exponent rules
af(x) · ag(x) = af(x) + g(x) ·
af(x)ag(x) = af(x) − g(x)
Combining logs
Log rules
log f(x) + log g(x) = log[f(x) · g(x)] ·
log f(x) − log g(x) = log[f(x)g(x)]
The hidden quadratic — substitution trick
Some equations are quadratics in disguise. The signature: three terms, where the variable in the first term is the square of the variable in the second:
Polynomial
x4 − 5x2 + 4 = 0
let u = x2
Exponential
e2x + ex − 6 = 0
let u = ex
Square root
x − 5√x + 6 = 0
let u = √x
🧭 Recipe — solving by substitution
- Spot the pattern: three terms, one is the square of another.
- Substitute u = the simpler term. The equation becomes a quadratic in u.
- Solve the quadratic using factoring, formula, or completing the square.
- Substitute back: replace u with the original expression and solve for x.
- Discard impossible solutions (e.g. if u = ex, then u must be positive — reject any negative u).
Don’t divide by an expression that could be zero
It’s tempting to cancel a common factor — but if that factor could equal zero, you lose solutions:
Lost solution example
(x − 3)(x + 1) = 4(x − 3)
✗ Dividing by (x − 3): x + 1 = 4 → x = 3 (just one solution — but wait, x = 3 also satisfies the original!)
The fix: move everything to one side and factor instead of dividing.
The safe move: rearrange to (stuff) = 0, factor, then set each factor to zero. You’ll catch every solution.
Worked examples
WE 1Solve a simple log equation
Find the exact solution of 7 − 3 log2 x = 1.
Step 1: Isolate the log term
−3 log₂ x = −6
log₂ x = 2
Step 2: Apply inverse — exponentiate base 2
x = 2² = 4
x = 4
always check: log₂ 4 = 2, then 7 − 3(2) = 1 ✓
WE 2Solve an equation involving a square root
Solve x = √(2x + 8).
Step 1: Square both sides — many-to-one, expect a check
x² = 2x + 8
x² − 2x − 8 = 0
Step 2: Factor
(x − 4)(x + 2) = 0
x = 4 or x = −2
Step 3: Check in the original
x = 4: LHS = 4, RHS = √16 = 4 ✓
x = −2: LHS = −2, RHS = √4 = 2 ✗
x = 4
squaring created a false solution at x = −2 — always check after squaring
WE 3Solve a hidden-quadratic exponential equation
Solve e2x − 7ex + 12 = 0.
Step 1: Notice e²ˣ = (eˣ)² — substitute u = eˣ
u² − 7u + 12 = 0
Step 2: Factor and solve
(u − 3)(u − 4) = 0
u = 3 or u = 4
Step 3: Substitute back u = eˣ
eˣ = 3 → x = ln 3
eˣ = 4 → x = ln 4
x = ln 3 or x = ln 4
both solutions are valid — eˣ is always positive, so both u-values are acceptable
WE 4Combine logs to solve a log equation
Find the exact solution of ln x + ln(x − 2) = ln 15.
Step 1: Combine the LHS using the log addition rule
ln[x(x − 2)] = ln 15
Step 2: Equate arguments (since ln is one-to-one)
x(x − 2) = 15
x² − 2x − 15 = 0
Step 3: Factor and solve
(x − 5)(x + 3) = 0
x = 5 or x = −3
Step 4: Check domain — log requires positive argument
x = 5: ln 5 ✓ and ln 3 ✓
x = −3: ln(−3) is undefined ✗
x = 5
log of a negative is undefined — reject any solution that breaks the log’s domain
WE 5Solve without dividing by a possibly-zero expression
Solve (x + 2)(x − 5) = 6(x + 2).
Step 1: DON’T divide by (x + 2). Move everything to one side
(x + 2)(x − 5) − 6(x + 2) = 0
Step 2: Factor out (x + 2)
(x + 2)[(x − 5) − 6] = 0
(x + 2)(x − 11) = 0
Step 3: Set each factor to zero
x + 2 = 0 → x = −2
x − 11 = 0 → x = 11
x = −2 or x = 11
had we divided by (x + 2), we’d have lost x = −2 — that’s why factoring is the safer move
WE 6Solve an exponential equation with different bases
Find the exact solution of 4x + 1 = 9x, giving your answer as a single logarithm.
Step 1: Take ln of both sides
(x + 1) ln 4 = x ln 9
Step 2: Expand and group x terms
x ln 4 + ln 4 = x ln 9
x ln 4 − x ln 9 = −ln 4
x(ln 4 − ln 9) = −ln 4
Step 3: Solve for x and tidy up
x = −ln 4 / (ln 4 − ln 9)
= ln 4 / (ln 9 − ln 4)
= ln 4 / ln(9/4)
x = ln 4 / ln(9/4)
“exact” answer means leave the logs in — don’t decimalise unless asked to
💡 Top tips
- Apply inverses to whole sides, not term by term. eln a + ln b = ab, NOT a + b.
- After squaring, always check. Squaring can manufacture false solutions.
- Combine log/exp terms before rearranging. Use log rules to get a single log on each side, then equate arguments.
- Spot hidden quadratics: three terms, one is the square of another → substitution time.
- Never divide by an expression that could be zero. Subtract and factor instead.
- For “exact” answers, leave logs/surds/fractions intact. Don’t reach for the calculator unless the question allows decimals.
- Check the domain of any log argument — log of a non-positive is undefined and means rejecting that solution.
⚠ Common mistakes
- Applying inverses term by term. eln x + ln(x−1) = e5 means x(x−1) = e5, NOT x + (x−1) = e5.
- Forgetting the ± after taking an even root. x2 = 49 has solutions x = 7 AND x = −7.
- Failing to check after squaring. Squaring can introduce extra solutions.
- Dividing by a factor that contains the unknown. You may lose a valid solution. Always factor instead.
- Accepting log solutions without checking domain. log of zero or negative is undefined.
- Mixing up log laws: log(a + b) ≠ log a + log b. (The right rule is log(ab) = log a + log b.)
- Decimalising “exact” answers. If the question wants exact, leave it as ln, log, fraction, or surd.
When algebra runs out — equations like ex = x2 mixing different function families — there’s no analytic method that works. The next note covers solving equations graphically: plotting both sides and reading off the intersection points using your GDC.
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