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IB Maths Paper 1 & 2 25 min read

Solving Exponential Equations

Hi! Now that you know your logarithms and laws of indices, it’s time to put them together. In this lesson, we’ll learn how to solve exponential equations — equations where the unknown x is sitting up in the power. There are 3 main techniques, and once you know which one to use for which type of question, every IB problem on this topic becomes a step-by-step recipe.

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What you will learn today

  • What an exponential equation is and how to spot one
  • Method 1: solve by inspection (looking for the answer)
  • Method 2: solve by changing the base (very common in Paper 1)
  • Method 3: solve using logarithms (when nothing else works)
  • How to handle quadratic-style exponential equations (the “hidden quadratic” trick)
  • 5 worked examples covering every type you’ll see in the exam

Step 1: What is an exponential equation?

An exponential equation is any equation where the unknown is in the power. Here are some examples:

Notice something? In every one of these, the x is in the exponent (the little number on top). That’s what makes them “exponential”. Compare this with a normal linear equation like 3x + 1 = 10, where x just sits on the line.

The big challenge: when x is in the power, you can’t just rearrange and divide like a normal equation. You need a special tool to “pull it down” first. That’s exactly what these 3 methods do.

Step 2: Method 1 — Solve by inspection

This is the easiest method, but it only works for simple equations where the answer is a small whole number. Basically, you look at the equation and try to spot the answer in your head.

METHOD 1

By Inspection (just see the answer)

Look for whole-number powers that match

When to use it: when both sides can be written as the same base, with simple whole-number powers.

Example 1:   3x = 9  →  you can see that 3² = 9, so x = 2.

Example 2:   2x = 64  →  2⁶ = 64, so x = 6.

Example 3:   5x = 125  →  125 = 15² = 5⁻², so x = −2.

Memory check before exam: know your powers of 2, 3, and 5 cold. Powers of 2: 2, 4, 8, 16, 32, 64, 128, 256. Powers of 3: 3, 9, 27, 81, 243. Powers of 5: 5, 25, 125, 625. If you know these instantly, half the exam questions become trivial.

Step 3: Method 2 — Solve by changing the base

This is the most common technique in Paper 1 (no calculator). The idea is simple: if both sides of the equation have different bases, rewrite them so the bases match. Then you can compare the powers directly.

METHOD 2

Change the Base (make both sides match)

If ap = aq,   then   p = q

When to use it: when both sides can be rewritten as powers of the same base. Look for numbers that are powers of each other (like 4 and 8 both being powers of 2).

The golden rule: once both sides have the same base, you can simply set the powers equal to each other and solve a normal equation.

Walkthrough — let’s solve 8x = 32

The bases 8 and 32 don’t match. But hold on — both are powers of 2! Let’s rewrite them.

  1. Start:   8x = 32
  2. Rewrite both sides as powers of 2.   8 = 2³ and 32 = 2⁵
  3. Substitute:   (2³)x = 2⁵
  4. Use the index law (power on power):   23x = 2⁵
  5. Now both sides have base 2. Set the powers equal:   3x = 5
  6. Solve:   x = 53

Perfect! By rewriting both sides with the same base, we turned a scary exponential equation into a basic linear one.

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Numbers worth memorising

To use change-of-base quickly, you need to spot when one number is a power of another. These come up over and over in IB exams:

Powers of 2:   4 = 2²  |  8 = 2³  |  16 = 2⁴  |  32 = 2⁵  |  64 = 2⁶

Powers of 3:   9 = 3²  |  27 = 3³  |  81 = 3⁴

Powers of 5:   25 = 5²  |  125 = 5³  |  625 = 5⁴

Step 4: Method 3 — Solve using logarithms

Sometimes the bases just can’t be matched. For example, in 2x = 7, there’s no nice way to write 7 as a power of 2. The answer won’t be a whole number — it’ll be something like 2.807… So we need a different tool: logarithms.

METHOD 3

Use Logarithms (when bases won’t match)

ax = b  ⇔  x = loga(b)

When to use it: when the answer is NOT a whole number, or when the bases can’t be matched. You’ll usually need a calculator for these.

Example:   2x = 7  →  x = log2(7) = 2.807…

The “take logs of both sides” trick

There’s another way to use logs that’s super helpful when the equation is more complicated. Instead of flipping into log form right away, you can take the log of both sides and then use the power law to bring the x down.

Let me show you. Solve 2x = 7 using this trick:

  1. Start:   2x = 7
  2. Take ln of both sides (you can use log base 10 too — same result):
        ln(2x) = ln(7)
  3. Use the power law of logs to bring the x down:
        x ln(2) = ln(7)
  4. Divide by ln(2):    x = ln(7)ln(2)
  5. Type into your GDC:   x = 2.807… ≈ 2.81 (3 s.f.)

Why bother taking logs of both sides? Because once you do, the power law of logs lets you pull x down from the exponent. Now x is just a regular variable you can solve for. This trick saves the day in tougher questions.

Step 5: How do I know which method to use?

Good question. Here’s a simple decision tree to help you pick the right method:

  1. Step A — Try inspection first. Can you see the answer just by looking? If yes, you’re done in 5 seconds.
  2. Step B — If no, can you match the bases? Look for numbers that are powers of each other (8 and 16 are both powers of 2, etc.). If yes, use Method 2 (change of base).
  3. Step C — If neither works, use logs. Method 3 is your fallback for any equation where the answer isn’t a whole number.

Use Inspection / Change of Base when…

The numbers are “nice” (powers of small bases like 2, 3, 5).

The expected answer is a fraction or whole number.

It’s a Paper 1 question (no calculator).

Use Logarithms when…

The bases just can’t be matched.

The expected answer is a decimal (e.g. 3 s.f.).

It’s a Paper 2 question (calculator allowed).

Step 6: The “hidden quadratic” trick

This is one of the most beautiful tricks in IB Maths, and it shows up in nearly every exam. Some exponential equations are actually quadratic equations in disguise!

Look at this:

(ax)² + b(ax) + c = 0

If we let u = ax, the equation becomes:

u² + bu + c = 0

That’s a normal quadratic! You can factorise or use the quadratic formula to find u, then go back and solve for x.

Watch out for the disguise! The “(ax)²” part is sometimes hidden as a2x. Remember the index law: a2x = (ax)². Once you spot this, you can substitute u = ax and the quadratic appears.

Example: solve 4x − 5(2x) + 4 = 0

  1. Spot the hidden quadratic. Notice 4x = (2²)x = (2x)². So the equation becomes:
        (2x)² − 5(2x) + 4 = 0
  2. Substitute u = 2x:
        u² − 5u + 4 = 0
  3. Factorise:   (u − 1)(u − 4) = 0
  4. So:   u = 1   or   u = 4
  5. Substitute back:   2x = 1   or   2x = 4
  6. Solve each by inspection:   x = 0   or   x = 2

Two solutions, both nice and clean. The hidden quadratic trick turned a confusing equation into one we already know how to handle.

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Reject negative solutions for u

When you let u = ax, remember that u must be positive (since ax is always positive for any real x). If your quadratic gives you a negative value of u, just throw that solution away — it doesn’t lead to a real answer for x.

Step 7: Worked examples

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Worked Example 1 — Solve by inspection

Solve:   5x = 125

Answer:

Step 1: think — what power of 5 gives 125? 5³ = 125 Step 2: so x = 3. x = 3 Inspection works perfectly here — no calculator needed.
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Worked Example 2 — Change of base

Solve:   3x + 1 = 19x

Answer:

Step 1: rewrite the right side using base 3. We know 9 = 3². 19x = 1(3²)x = 132x = 3−2x Step 2: equation now becomes: 3x + 1 = 3−2x Step 3: same base → set powers equal. x + 1 = −2x Step 4: solve for x. 3x = −1 x = −13
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Worked Example 3 — Use logarithms

Solve:   5x = 8, giving your answer to 3 s.f.

Answer:

Step 1: this can’t be solved by inspection (5¹ = 5 too small, 5² = 25 too big), so use logs. Step 2: take ln of both sides. ln(5x) = ln(8) Step 3: use the power law to bring x down. x ln(5) = ln(8) Step 4: divide by ln(5). x = ln(8)ln(5) Step 5: type into your GDC. x = 1.292… x = 1.29 (3 s.f.)
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Worked Example 4 — Hidden quadratic (exam-style)

Solve:   4x − 3(2x + 1) + 9 = 0. Give your answer correct to 3 significant figures.

Answer:

Step 1: spot the hidden quadratic. 4x = (2²)x = (2x)² Step 2: also expand 2x + 1 using index laws. 2x + 1 = 2x × 2¹ = 2 × 2x Step 3: rewrite the equation. (2x)² − 3(2 × 2x) + 9 = 0 (2x)² − 6(2x) + 9 = 0 Step 4: substitute u = 2x. u² − 6u + 9 = 0 Step 5: factorise (it’s a perfect square!). (u − 3)² = 0  →  u = 3 Step 6: substitute back: 2x = 3. Step 7: take logs to solve. x ln(2) = ln(3) x = ln(3)ln(2) = 1.584… x = 1.58 (3 s.f.)
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Worked Example 5 — Two solutions

Solve:   9x − 4(3x) + 3 = 0

Answer:

Step 1: spot that 9x = (3²)x = (3x)². (3x)² − 4(3x) + 3 = 0 Step 2: substitute u = 3x. u² − 4u + 3 = 0 Step 3: factorise. (u − 1)(u − 3) = 0 u = 1   or   u = 3 Step 4: solve each by inspection. 3x = 1  →  x = 0 3x = 3  →  x = 1 x = 0   or   x = 1 Both u values were positive — both gave valid answers!
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Examiner Tips

Read the question carefully. If it asks for an “exact” answer, leave it in log form like ln(7)ln(2). If it asks for “3 s.f.” or “decimal”, evaluate using your GDC.

Always check for hidden quadratics. Whenever you see something like 4x AND 2x in the same equation, or 9x AND 3x, that’s your hint — it’s a quadratic in disguise.

Don’t forget to substitute back. If you let u = 2x, the answer to the quadratic gives you u — not x! You still need to solve 2x = u to find x.

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Mistakes to avoid

  • Forgetting to substitute back. If u = 2x and you find u = 8, that’s NOT the final answer. You still need to solve 2x = 8 → x = 3.
  • Missing the hidden quadratic. Always check if you can write one term as the square of another (4x = (2x)²).
  • Accepting negative u values. If u = ax, then u > 0 always. A negative u means no real solution — discard it.
  • Confusing 2x + 1 with 2x + 1. Big difference! 2x + 1 = 2 × 2x (the +1 is part of the exponent). Don’t drop the bracket in your head.
  • Mixing up base when changing. When rewriting 8 as 2³, then raising to x: (2³)x = 23x, NOT 2³x. The whole exponent is multiplied.
  • Not knowing your small powers. If you don’t have powers of 2, 3, 5 memorised, you’ll waste exam time figuring out 2⁵ = 32 or 3⁴ = 81. Memorise them!
  • Using a calculator on Paper 1. No calculator! On Paper 1, every answer must come out as a clean number using inspection or change of base. If you find yourself wanting a calculator, you’ve probably missed a base-matching trick.

Final word from your teacher: exponential equations look scary because x is “stuck up there” in the power, but the 3 methods cover every possible question. Always start with inspection (it’s free!), then try changing the base, and only use logs as a last resort. For tougher questions, look out for the hidden quadratic — it’s the IB’s favourite trick. Practise 15–20 mixed problems and you’ll start to recognise the pattern instantly.

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