IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 ~10 min read

Solving Exponential Equations

An exponential equation is one where the unknown sits in the exponent — like 3^x = 81 or 5 · 2^{x − 1} = 40. The earlier notes gave you the engine (the flip identity) and the gears (the log laws). Now you put them to work. Three solving methods cover almost every exam question — pick the right one and the rest is mechanical. The trickier questions hide a quadratic in disguise; spot it and you’re done.

📘 What you need to know

An exponential equation has the unknown in the power — for example a^x = b or k · a^f(x) = m.
Three solving methods, in order of preference: inspection, same-base equating, logarithms.
Inspection works for clean cases like 2^x = 32.
Same-base works when both sides can be written as powers of the same prime (often 2, 3, or 5). Once same-base, set the exponents equal.
Logs are the catch-all — take ln (or log) of both sides and use the power law to bring the exponent down.
Look for hidden quadratics: if you see a^2x alongside a^x, substitute u = a^x and you’ll have a normal quadratic in u.
If the question says “exact“, leave the answer as a logarithm. If it says “3 s.f.“, finish on the GDC.

Picking the right method

Before solving anything, ask three questions in order. Stop at the first “yes”.

Which method should I use?

Method 1 — Inspection

If the answer is obvious, just write it down. Many exam questions either start or end with a step that needs this — recognising small powers without the GDC is a skill the IB rewards directly.

Worth memorising: powers of 2 up to 2¹⁰ (= 1024) | powers of 3 up to 3⁵ (= 243) | powers of 5 up to 5⁴ (= 625) | powers of 10 up to 10⁶. Most “spot the answer” exponents come from this list.

Method 2 — Equate exponents (same base)

If both sides of the equation can be written as powers of the same base, the equation reduces to a simple polynomial equation in the exponents.

Same-base principle a^f(x) = a^g(x) ⇔ f(x) = g(x)

🧭 Recipe — equate exponents

Spot a common base. Often 2, 3, or 5. Numbers like 4, 8, 16 are powers of 2. Numbers like 9, 27, 81 are powers of 3.
Rewrite each term using that base, applying the index laws as needed.
Equate the exponents — set the two power expressions equal.
Solve the resulting equation in x (often linear, sometimes quadratic).

Useful disguises: a^2x = (a^x)² | a^{x + 1} = a · a^x | a^−x = 1a^x. The last two come straight from the index laws.

Method 3 — Logarithms (the catch-all)

When neither inspection nor same-base works, take logs of both sides. Any base will do, but most students use ln (because the GDC has a button) or log_a (when the base of the equation is convenient).

🧭 Recipe — solve with logs

Isolate the exponential term on one side, if it’s not already (move any constants over first).
Take the natural log (or any log) of both sides.
Use the power law to bring the exponent down as a coefficient: ln(a^f(x)) = f(x) · ln(a).
Rearrange and solve for x.

If the question asks for an exact answer, you’ll usually leave it as a logarithm — for example x = ln(20) / ln(3). Don’t punch it into the GDC unless the question wants a decimal.

Hidden quadratics

This is the only “trick” type at HL — and it shows up frequently. The equation looks exponential, but if you stare at it long enough, you spot a quadratic shape underneath.

The hidden form A(a^x)² + B(a^x) + C = 0 → let u = a^x ✗ not a formula — it’s a substitution technique

🤔 How to spot the quadratic

If the equation has both an a^2x term and an a^x term, the a^2x rewrites as (a^x)². The whole equation is then a quadratic in a^x. Substitute u = a^x, factor, solve the quadratic, then back-substitute to recover x.

Common giveaways: 4^x with 2^x | 9^x with 3^x | e^2x with e^x. Whenever you see one with the other, suspect a hidden quadratic.

After back-substituting, you’ll usually have one or two equations of the form a^x = (some number). Solve each one with whichever of the three methods is fastest. Reject any branch where a^x = (negative) — exponentials are always positive, so that branch has no solution.

Worked examples

WE 1

Solve by inspection

Solve 2^x = 64.

Read it as a question — what power of 2 gives 64? 2¹ = 2, 2² = 4, …, 2⁶ = 64 ✓ x = 6

WE 2

Same base on both sides

Solve 9^x = 27^{x − 1}.

Step 1: Both 9 and 27 are powers of 3 9 = 3², 27 = 3³ Step 2: Rewrite both sides (3²)^x = (3³)^{x − 1} 3^2x = 3^{3x − 3} Step 3: Equate the exponents 2x = 3x − 3 x = 3 no logs needed — same-base equating wins when the numbers cooperate

WE 3

Solve using logs — when no shortcut works

Solve 3^x = 20, giving your answer to 3 s.f.

Step 1: 20 isn’t a power of 3 — go straight to logs Step 2: Take ln of both sides ln(3^x) = ln(20) Step 3: Bring the exponent down with the power law x · ln(3) = ln(20) x = ln(20)ln(3) Step 4: GDC x = 2.73 (3 s.f.) exact answer would be ln(20)/ln(3) — leave it like that if “exact” is asked for

WE 4

Equation with base e

Solve e^2x = 17, giving your answer in exact form.

Step 1: Take ln of both sides ln(e^2x) = ln(17) Step 2: ln(e^k) = k 2x = ln(17) x = ln(17)2 when the base is e, ln cancels it cleanly — much faster than ln-then-divide

WE 5

Hidden quadratic — integer solutions

Solve 4^x − 5(2^x) + 4 = 0.

Step 1: Spot the hidden quadratic — 4^x = (2^x)² (2^x)² − 5(2^x) + 4 = 0 Step 2: Substitute u = 2^x u² − 5u + 4 = 0 (u − 1)(u − 4) = 0 u = 1 or u = 4 Step 3: Back-substitute 2^x = 1 → x = 0 2^x = 4 → x = 2 x = 0 or x = 2 two valid solutions — both kept since 2ˣ is positive in each case

WE 6

Hidden quadratic — exact log solutions

Solve e^2x − 5e^x + 6 = 0, giving your answers in exact form.

Step 1: e^2x = (e^x)² (e^x)² − 5(e^x) + 6 = 0 Step 2: Let u = e^x u² − 5u + 6 = 0 (u − 2)(u − 3) = 0 u = 2 or u = 3 Step 3: Back-substitute and take ln e^x = 2 → x = ln(2) e^x = 3 → x = ln(3) x = ln(2) or x = ln(3) both u-values are positive, so both branches give valid solutions

💡 Top tips

Try inspection first. If the answer is small and obvious, write it down — don’t burn time setting up logs.
Look for a common base before reaching for logs. If both sides factor as powers of 2 (or 3, or 5), you can skip logs entirely.
For hidden quadratics, the dead giveaway is seeing a^2x alongside a^x. Substitute u, factor, back-sub.
Reject negative u-values: a^x > 0 always. If the quadratic gives u = −2, that branch has no solutions — discard it.
If the equation has base e, take ln (not log). Saves a step.
Read the answer style carefully: “exact” means leave logs; “3 s.f.” means decimal.
For a^{x + k}, factor the constant out: a^{x + k} = a^k · a^x. This often turns a messy equation into a hidden quadratic.

⚠ Common mistakes

“Cancelling” different bases. 2^x = 3^{x − 1} does not give x = x − 1. Different bases — must take logs.
Keeping a negative u branch. If u = a^x = −5 falls out of the quadratic, that branch has no real solutions. Reject it.
Forgetting the power law. ln(3^x) = x · ln(3), not ln(3) · x^… or anything else. The exponent comes down.
Missing the “+1” trick. 2^{x + 1} means 2 × 2^x, not (2^x)². Index laws — not power-of-a-power.
Skipping the back-substitution after solving for u. The question is asking for x, not u. You must convert.
Using the wrong base when taking logs. Both sides need the same log. Mixing ln on one side with log₁₀ on the other gives nonsense.
Confusing a^x with x^a. Exponential equations have x in the power. x² = 9 is a polynomial equation — different toolkit entirely.

Exponential equations are a routine type once the toolkit is in place. Practice spotting which method is fastest — this single judgement call saves more exam time than any single technique. The same skills carry forward into modelling, growth/decay, and continuous compound interest later in the course.

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