Applications of Logarithmic Functions
Key Terms
- Solving loga(x) = b
- convert to exponential form → x = ab.
- Solving loga(f(x)) = loga(g(x))
- equate arguments → f(x) = g(x) (then check domain).
- Using log laws first
- simplify both sides to a single log before solving.
- Domain restriction
- always check that all arguments are positive in the original equation. Discard any solution that makes an argument ≤ 0.
- pH
- pH = −log[H+], so [H+] = 10−pH.
- Richter scale
- M = log(I / I0), so I = I0 × 10M.
- Population / exponential growth
- P = P0 ekt; solve for t using ln: t = ln(P / P0) / k.
Strategy 1: loga(x) = b → x = ab (convert to exponential form)
Strategy 2: loga(f(x)) = loga(g(x)) → f(x) = g(x) (equate arguments — same base required)
Strategy 3: Use log laws to combine multiple logs into one, then apply Strategy 1 or 2.
| Model | Formula | Solve for unknown |
|---|---|---|
| pH | pH = −log[H+] | [H+] = 10−pH |
| Richter scale | M = log(I / I0) | I = I0 × 10M |
| Population growth | P = P0 ekt | t = ln(P / P0) / k |
| Compound interest | A = P(1 + r)t | t = log(A / P) / log(1 + r) |
Worked Example 1 — Solving a basic logarithmic equation
Solve log3(2x − 1) = 4.
Step 1: Convert to exponential form: 2x − 1 = 34 = 81
Step 2: Solve: 2x = 82 → x = 41
Check domain: 2(41) − 1 = 81 > 0 ✓
Worked Example 2 — Equations with logs on both sides
Solve log2(x + 3) = log2(2x − 1).
Step 1: Same base → equate arguments: x + 3 = 2x − 1
Step 2: Solve: 3 + 1 = 2x − x → x = 4
Check domain: x + 3 = 7 > 0 ✓ and 2(4) − 1 = 7 > 0 ✓
Worked Example 3 — Using log laws before solving
Solve log(x) + log(x − 3) = 1.
Step 1: Combine using product law: log(x(x − 3)) = 1
Step 2: Convert: x(x − 3) = 101 = 10
Step 3: Expand and solve: x2 − 3x − 10 = 0 → (x − 5)(x + 2) = 0 → x = 5 or x = −2
Check domain: Need x > 0 and x − 3 > 0, so x > 3. x = 5: 5 > 3 ✓ x = −2: fails. x = 5 only.
Worked Example 4 — Application: population growth
A population grows according to P = 500e0.04t, where t is in years. After how many years does the population reach 2 000?
Step 1: Set P = 2000: 500e0.04t = 2000 → e0.04t = 4
Step 2: Take natural log of both sides: 0.04t = ln(4)
Step 3: Solve: t = ln(4) / 0.04 = 1.3863 / 0.04 ≈ 34.7 years
Introduction
Logarithms appear throughout science, finance, and engineering because many real-world quantities grow or decay exponentially, and taking a logarithm is the natural way to “undo” that growth. This lesson develops three core solving strategies and then applies them to models from chemistry, seismology, biology, and finance.
Strategy 1: Convert to Exponential Form
The definition of a logarithm gives us a direct way to solve equations of the form loga(expression) = number:
loga(x) = b ⇔ x = ab
This is the most important single technique. Once you convert, you have a straightforward algebraic equation to solve.
Step 1: Identify a = 4, b = 3, and the argument is x + 2.
Step 2: Convert: x + 2 = 43 = 64.
Step 3: Solve: x = 62.
Step 4: Check: argument = 62 + 2 = 64 > 0 ✓. Answer: x = 62.
Step 1: Isolate the log: 2log(x) = 4 → log(x) = 2.
Step 2: Convert (base 10): x = 102 = 100.
Step 3: Check: 100 > 0 ✓. Answer: x = 100.
Strategy 2: Equate Arguments (Same Base)
If the equation has the form loga(f(x)) = loga(g(x)) — the same base and a single log on each side — then the arguments must be equal:
loga(f(x)) = loga(g(x)) ⇔ f(x) = g(x)
This only works when both sides have exactly one log with the same base. Always check both arguments are positive at the solution.
Step 1: Same base → equate: 3x + 1 = x + 7.
Step 2: Solve: 2x = 6 → x = 3.
Step 3: Check: 3(3)+1=10>0 ✓ and 3+7=10>0 ✓. Answer: x = 3.
Strategy 3: Simplify Using Log Laws, then Solve
When an equation contains multiple logs on one side (e.g. log(x) + log(x+2)), use log laws to combine them into a single logarithm first, then apply Strategy 1 or 2. This is where domain checks are most critical, because factoring often produces an extraneous (invalid) solution.
Step 1: Use quotient law: log2((x + 4) / (x − 1)) = 3.
Step 2: Convert: (x + 4) / (x − 1) = 23 = 8.
Step 3: Solve: x + 4 = 8(x − 1) = 8x − 8 → 12 = 7x → x = 12/7.
Step 4: Check: x+4 = 12/7+28/7 = 40/7 > 0 ✓. x−1 = 12/7−7/7 = 5/7 > 0 ✓. Answer: x = 12/7.
Real-World Applications
Logarithmic equations arise in many real contexts. In each case, the same three strategies apply — the only difference is recognising which model to use.
The pH of a solution measures its acidity: pH = −log[H+], where [H+] is the hydrogen ion concentration in mol/L. To find [H+] from pH: [H+] = 10−pH. A difference of 1 pH unit corresponds to a 10× change in [H+].
Example: Find the pH of a solution with [H+] = 3.5 × 10−4 mol/L.
pH = −log(3.5 × 10−4) = −[log(3.5) + log(10−4)] = −[0.544 − 4] = −(−3.456) ≈ 3.46.
Earthquake magnitude: M = log(I / I0), where I is the earthquake's intensity and I0 is the reference intensity. An earthquake of magnitude 6 is 10 times more intense than one of magnitude 5.
Example: How much more intense is a magnitude 7 earthquake than a magnitude 4?
Ratio = 107 / 104 = 103 = 1 000 times more intense.
Continuous growth: P = P0 ekt, where P0 is the initial value, k is the growth rate, and t is time. To solve for t:
P / P0 = ekt → ln(P / P0) = kt → t = ln(P / P0) / k
For compound interest with annual compounding: A = P(1 + r)t → t = log(A/P) / log(1 + r).
Summary
To solve any logarithmic equation: (1) identify the structure, (2) apply the appropriate strategy — convert to exponential form, equate arguments, or use log laws first, (3) solve the resulting algebraic equation, and (4) check all solutions satisfy the domain restrictions of the original equation.
Mastery Practice
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Fluency
Solve each equation. State any domain restrictions and check your solution.
- (a) log2(x) = 6
- (b) log5(x − 3) = 2
- (c) log(x + 1) = 3
- (d) ln(x) = 4
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Fluency
Solve each equation (logs on both sides). State any domain restrictions.
- (a) log3(x + 5) = log3(2x − 1)
- (b) log(4x − 3) = log(2x + 7)
- (c) ln(3x) = ln(x + 8)
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Fluency
pH calculations.
- (a) Find the pH of a solution with [H+] = 10−5 mol/L.
- (b) Find the pH of a solution with [H+] = 2.5 × 10−3 mol/L. Give your answer to 2 decimal places.
- (c) Find [H+] for a solution with pH = 8.3. Give your answer in scientific notation to 2 significant figures.
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Fluency
Solve using log laws first. Show all working and check for extraneous solutions.
- (a) log2(x) + log2(x − 2) = 3
- (b) log(x + 4) − log(x − 1) = 1
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Understanding
Richter scale questions.
The Richter scale magnitude is M = log(I / I0), where I0 is the reference intensity.
- (a) The 2011 Tōhoku earthquake had magnitude M = 9.0. Find its intensity as a multiple of I0.
- (b) A minor tremor has M = 3.2. How many times more intense is the Tōhoku earthquake (M = 9.0) compared to this tremor?
- (c) Two earthquakes have magnitudes M1 and M2. Show that the ratio of their intensities is 10M1 − M2.
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Understanding
Exponential equations solved using logarithms.
- (a) Solve 5x = 80. Give your answer to 3 decimal places.
- (b) Solve 3 × 2x = 48.
- (c) Solve e2x − 1 = 7. Give your answer in exact form and as a decimal (3 d.p.).
-
Understanding
Population growth model.
The population of a town grows according to P = 12 000 e0.025t, where t is the number of years after 2020.
- (a) What is the population in 2020? What does the constant 0.025 represent?
- (b) Find the population in 2030 (t = 10). Give your answer to the nearest whole number.
- (c) In what year does the population first exceed 20 000? Show full working.
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Understanding
Solve each equation. Show all working.
- (a) log3(x + 1) + log3(x + 3) = log3(5)
- (b) log2(x − 1) − log2(x + 2) = −2
- (c) 2log5(x) − log5(x − 4) = 2
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Problem Solving
Compound interest problem.
Challenge. Rachel invests $8 000 in an account paying 4.5% per annum compound interest.- (a) Write an expression for the value A of the investment after t years.
- (b) Use logarithms to find, to the nearest month, how long it takes for the investment to double.
- (c) How long does it take for the original investment to triple? Give your answer to the nearest year.
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Problem Solving
Comparing two models.
Challenge. Two radioactive substances A and B have masses (in grams) at time t (in years) given by:
MA = 200 e−0.06t and MB = 150 e−0.04t- (a) Find the initial mass of each substance.
- (b) At what time do the two substances have the same mass? Give your answer to 2 decimal places.
- (c) After 50 years, which substance has more mass? Justify your answer with calculation.