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. e^2x + e^x + 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. e^{ln 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 = e^x ⟺ x = ln y

take ln to bring down the exponent

log ↔ exp

y = log_a x ⟺ x = a^y

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 a^f(x) · a^g(x) = a^{f(x) + g(x)} · a^f(x)a^g(x) = a^{f(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

e^2x + e^x − 6 = 0

let u = e^x

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 = e^x, 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 1

Solve 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 2

Solve 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 3

Solve a hidden-quadratic exponential equation

Solve e^2x − 7e^x + 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 4

Combine 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 5

Solve 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 6

Solve an exponential equation with different bases

Find the exact solution of 4^{x + 1} = 9^x, 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. e^{ln 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. e^{ln 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 e^x = 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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