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NotesMath AI HLTopic 1.5Solving Exponential and Logarithmic Equations
Back to Math AI HL Topics
1.5.42 min read

Solving Exponential and Logarithmic Equations

IB Mathematics: Applications and Interpretation • Unit 1

Exam preparation

Practice the questions examiners actually ask

Our question bank mirrors real IB exam papers. Practice under timed conditions and track your progress across topics.

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Contents

  • Solving exponential equations by common base
  • Solving exponential equations with logarithms
  • Solving logarithmic equations
  • Checking solutions and domain sense
  • IB-style model questions and exam strategy
The big idea: If both sides can be written with the same base, then the exponents must be equal.

Worked example

Solve 2x+1 = 16.

Step by step

  1. Rewrite 16 as a power of 2.
  2. Rewrite the equation with the common base.
  3. Since the bases are equal, the exponents must be equal too — drop the common base.
  4. Solve.

Final answer

x = 3

Use common bases when you can: This is often the fastest method and avoids unnecessary calculator work.
When bases do not match: If you cannot rewrite both sides with a common base, use logarithms to bring the exponent down.

Worked example

Solve 3x = 20.

Step by step

  1. Take logs of both sides.
  2. Use the power law.
  3. Solve for x.

Final answer

x ≈ 2.727

Any consistent log base works: You can use log or ln, as long as you use the same one on both sides.

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The big idea: A logarithmic equation can often be rewritten as an exponential equation.

Worked example

Solve log2 x = 5.

Step by step

  1. Rewrite in exponential form.
  2. Calculate.

Final answer

x = 32

Second example

Solve log10 x = 2.3.

Step by step

  1. Rewrite in exponential form.
  2. Use a calculator.

Final answer

x ≈ 199.53

Logs need positive inputs: log x only makes sense when x > 0 — you cannot take the logarithm of 0 or a negative number in the real numbers.



So if you solve a log equation and get something like x = −100, reject it: log(−100) is undefined.



👉 Always check your answer back in the original equation — negative or zero answers should be rejected when the unknown sits inside a log.

🎯 IB-style worked example

Two GDC keys you'll use here:
  • LOG → use for part (a) when concentration is given and you want pH.
  • 2nd + LOG → 10ˣ; use for parts (b) and (c) to undo the log and find [H⁺].
Scenario: The pH of a solution tells you how acidic it is. The smaller the pH, the stronger the acid.



pH is linked to the hydrogen ion concentration [H⁺] (measured in moles per litre) by the formula:



pH = −log₁₀[H⁺]



Examples: pH 2 (lemon juice) is much more acidic than pH 5 (black coffee), which is more acidic than pH 7 (water).

Part (a) — find pH from concentration

Using pH = −log₁₀[H⁺]:



(a) Find the pH when [H⁺] = 0.00042 mol L⁻¹. [2 marks]

Step by step

  1. Substitute [H⁺] = 0.00042 into the formula:
  2. Evaluate log(0.00042) on the GDC. The minus sign in the formula flips the negative log into a positive pH:
  3. Round to 3 s.f.:

Final answer

pH ≈ 3.38


Part (b) — find concentration from pH

Using the same formula:



(b) Find [H⁺] when pH = 6.8. Give your answer in the form a × 10ᵏ. [2 marks]

Step by step

  1. Substitute pH = 6.8 into the formula:
  2. Multiply both sides by −1 to isolate the log:
  3. Undo the log with 10ˣ:

Final answer

[H⁺] ≈ 1.58 × 10⁻⁷ mol L⁻¹


Part (c) — compare two acids

A juice has pH 2.3 and tea has pH 4.9.



(c) How many times more acidic is the juice than the tea? [3 marks]

Step by step

  1. Convert both pH values to [H⁺] using 10ˣ:
  2. Divide stronger by weaker to get the ratio. Use the index law :
  3. Evaluate on the GDC:
  4. Write the answer in context:

Final answer

Juice is ≈ 398 times more acidic than tea.

Exam tips for pH questions:
  • The minus sign matters. pH = −log[H⁺] gives a positive pH because [H⁺] is small (less than 1), so log[H⁺] is negative.
  • Use the (−) key, not the subtract key, for negative exponents like 10−6.8.
  • For 'how many times stronger' questions, divide the larger concentration by the smaller — the ratio comes out clean using the index law 10a ÷ 10b = 10a−b.
  • Always keep units — concentration is in mol L⁻¹ (moles per litre).

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What you'll practise:
  • Plug numbers into a log model to find a missing constant.
  • Substitute a data pair to find another constant.
  • Find the smallest and biggest possible value of N.
  • Turn a rate into a waiting time using 1/N.
Two GDC keys you'll use:
  • LOG → evaluates log₁₀ (use for part (a) to find a).
  • 2nd + LOG → 10ˣ (use for part (b) to convert a into b).

✏️ IB-style worked example

Scenario: Background: Earthquake size is described by a magnitude M on the Richter scale, which goes from about 0 (barely detectable) up to roughly 8 (catastrophic). Stronger quakes are much rarer than weaker ones.



For a given region, the average number of earthquakes per year, N, with magnitude at least M is modelled by log₁₀ N = a − M for some constant a.



Suppose that in one region there are, on average, 50 earthquakes per year with magnitude at least M = 3.

Part (a) — find a

Using log₁₀ N = a − M with N = 50 and M = 3:



(a) Find a. [2 marks]

Step by step

  1. Substitute N = 50 and M = 3 into the model:
  2. Solve for a by adding 3 to both sides:
  3. Type into the GDC with LOG, then round to 3 s.f.:

Final answer

a ≈ 4.70


Part (b) — rewrite as N = b / 10M and find b

Using the same model:



(b) Show that the model can be written as N = b / 10^M and find b. [2 marks]

Step by step

  1. The question gives you the target form N = b / 10M and the data pair N = 50 when M = 3. So just substitute straight in:
  2. Multiply both sides by 10³ to isolate b:
  3. Write in standard form (the answer to 3 s.f.):

Final answer

b = 5.0 × 10⁴


Part (c) — range of N when 0 < M < 8

(c) For 0 < M < 8 (covering the full Richter range from barely felt to catastrophic), find the range of N. [2 marks]



Use N = b / 10^M with b = 5.0 × 10⁴ (= 50000).

Step by step

  1. Pick the two endpoints of M. The Richter scale runs from 0 to 8, so try M just above 0 and M just below 8. Use the model from part (b):
  2. Endpoint M = 0. When M is tiny, 10M is close to 1, so N is largest:
  3. Endpoint M = 8. When M is big, 10M is huge, so N is smallest:
  4. Write the range. N can be anywhere between these two values (use strict < because M is strictly between 0 and 8):

Final answer

0.0005 < N < 50000


Part (d) — expected waiting time for M ≥ 5.8

Using the same model:



(d) The largest recent earthquake had M = 5.8. The expected waiting time (in years) for the next earthquake of at least this size is 1/N. Find this time to the nearest year. [2 marks]

Step by step

  1. Recall the model from part (b) with b = 50000:
  2. Substitute M = 5.8 into the model:
  3. Take the reciprocal to get the waiting time in years:
  4. Round to the nearest year:

Final answer

≈ 13 years

Exam tips for log-model questions:
  • Find 'a' first — every other part needs it.
  • Keep a unrounded. Use Ans on the GDC to avoid rounding errors between parts.
  • N decreases as M increases. Plug in the endpoints to get the range.
  • Waiting time = 1/N. Always take the reciprocal after you find N — never invert M.

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3x+1 = 81. [2 marks]

Related Math AI HL Topics

Continue learning with these related topics from the same unit:

1.1.1Converting to standard form
1.1.2Back to ordinary form
1.1.3Calculations with standard form
1.1.4Validity checks and GDC output
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