Logarithm Word Problems

Use logarithms to solve real problems: pH, decibels, earthquake magnitudes, doubling time, half-life and how long money takes to grow.

Advanced35 min18 problems
Practice Problems

Definition

A logarithm word problem describes a real situation in which the unknown is an exponent: a time, a magnitude or a level. You solve it in three moves: write the formula, isolate the power, then take a logarithm of both sides.
Two kinds of problems appear again and again:
1. Logarithmic scales (the formula already contains a logarithm):
  • pH of a solution: pH=−log⁡[H+]\text{pH} = -\log[H^+], where [H+][H^+] is the hydrogen ion concentration in mol/L
  • Sound level in decibels: L=10log⁡(II0)L = 10 \log\left(\frac{I}{I_0}\right), where I0I_0 is the threshold of hearing
  • Richter magnitude: two earthquakes whose magnitudes differ by dd have amplitudes in the ratio 10d10^{d}
2. Growth and decay over time (the unknown tt sits in the exponent):
A=P(1+r)t⇒t=ln⁡(AP)ln⁡(1+r)A = P(1 + r)^t \quad\Rightarrow\quad t = \frac{\ln\left(\frac{A}{P}\right)}{\ln(1 + r)}
For continuous growth A=PertA = Pe^{rt} gives t=ln⁡(A/P)rt = \frac{\ln(A/P)}{r}, and for a half-life hh the amount N=N0(12)t/hN = N_0 \left(\frac{1}{2}\right)^{t/h} gives t=h⋅ln⁡(N/N0)ln⁡(0.5)t = h \cdot \frac{\ln(N/N_0)}{\ln(0.5)}.
Here log⁡\log means the common logarithm (base 10) and ln⁡\ln the natural logarithm (base ee). Most answers are decimals, so every problem states how to round, and you round only at the very end.

Try it now

An earthquake of magnitude 66 is compared with one of magnitude 44 on the Richter scale. How many times larger is the amplitude of the stronger earthquake?

Worked Examples

A soft drink has a hydrogen ion concentration of [H+]=3.2×10−4[H^+] = 3.2 \times 10^{-4} mol/L. Find its pH, rounded to two decimal places.
1

Write the formula

pH=−log⁡[H+]=−log⁡(0.00032)\text{pH} = -\log[H^+] = -\log(0.00032) → pH=−log⁡(0.00032)\text{pH} = -\log(0.00032)
2

Evaluate the common logarithm

log⁡(0.00032)≈−3.49485\log(0.00032) \approx -3.49485
3

Apply the minus sign

pH=−(−3.49485)=3.49485\text{pH} = -(-3.49485) = 3.49485
4

Round to two decimal places

The third decimal is 44, so round down → pH≈3.49\text{pH} \approx 3.49

Common Mistakes

Saying a magnitude 8 earthquake is "2 times" or "twice" as strong as a magnitude 6 earthquake

Why it's wrong: The Richter scale is logarithmic: each whole step multiplies the amplitude by 10. Differences in magnitude are exponents, not multipliers.

Correct: Ratio of amplitudes =108−6=102=100= 10^{8 - 6} = 10^2 = 100 times.

Forgetting the minus sign in pH=−log⁡[H+]\text{pH} = -\log[H^+] and getting a negative pH like −3.49-3.49

Why it's wrong: Concentrations such as 0.000320.00032 are less than 1, so their logarithm is negative. The minus sign in the formula makes the pH positive.

Correct: pH=−log⁡(0.00032)=−(−3.49485)≈3.49\text{pH} = -\log(0.00032) = -(-3.49485) \approx 3.49

Using the rate instead of the growth factor: t=ln⁡2ln⁡0.05t = \frac{\ln 2}{\ln 0.05}

Why it's wrong: The balance is multiplied by 1+r=1.051 + r = 1.05 each year, not by 0.050.05. Also ln⁡0.05\ln 0.05 is negative, which would give a negative time.

Correct: t=ln⁡2ln⁡(1.05)≈14.21t = \frac{\ln 2}{\ln(1.05)} \approx 14.21 years. For decay use 1−r1 - r, for example 0.850.85 for a 15%15\% loss.

Rounding too early, for example writing ln⁡1.05≈0.05\ln 1.05 \approx 0.05

Why it's wrong: Small roundings inside a quotient grow: 0.69310.05≈13.86\frac{0.6931}{0.05} \approx 13.86 instead of 14.2114.21, which is wrong by about 4 months.

Correct: Keep full calculator precision (or at least 5 significant digits) and round only the final answer, as the problem says.

Interactive Visual

Function graph

05101520253001000200030004000YearsDollars(14.21, 2000)

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  • Intersection point: (14.21, 2000)
10³ = 1000
3

See how powers of a number grow on the number line. Change the base and exponent.

10³ = 1000
3

See how powers of a number grow on the number line. Change the base and exponent.

Interest Calculator

Formula
Compound Interest:A = P(1 + r/n)^(nt)A = 1000 · (1 + 0.05)^15 = $2,078.93

Compound Interest

Interest Earned:$1,078.93
Total Amount:$2,078.93

Growth Over Time

yearsAmount

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051015202530-10-50510xy

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m=2, b=1

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Practice Problems

18 problems
Problem 1 of 18
Easy
An earthquake of magnitude 66 is compared with one of magnitude 44 on the Richter scale. How many times larger is the amplitude of the stronger earthquake?

Why It Matters

Logarithms turn huge ranges of numbers into small, readable ones, and they are the only tool that pulls an unknown out of an exponent:
  • Chemistry: lemon juice (pH about 2) has 100 000100\,000 times more hydrogen ions than pure water (pH 7).
  • Health: a 110 dB concert is 10001000 times more intense than an 80 dB street, which is why hearing protection matters.
  • Geology: a magnitude 7 earthquake shakes with 100100 times the amplitude of a magnitude 5 one.
  • Money: banks, savers and loan officers ask "how many years until my money doubles?" and answer it with ln⁡\ln.
  • Medicine: doctors use half-life to decide when a medicine has dropped below an effective level.
Every one of these questions is an exponent question, and a logarithm answers it.

Real World Applications

Acid Rain

Environmental scientists compare the acidity of rain with pH. Because pH is a logarithm, a small pH difference means a big difference in hydrogen ions.

Example:

Normal rain has a pH of about 5.65.6. Rain with pH 4.64.6 has 101=1010^{1} = 10 times as many hydrogen ions.

1Try It Yourself

Normal rain has a pH of 5.65.6. A storm brings acid rain with a pH of 4.34.3.

How many times more hydrogen ions does the acid rain contain? Round to the nearest whole number.

Step 1: Write the mathematical expression

Compute 105.6−4.310^{5.6 - 4.3} and round to the nearest whole number:

Hearing Protection at Concerts

Sound engineers and audiologists work in decibels. Ear plugs are rated by how much they cut the intensity of sound.

Example:

Dividing the intensity by 1010 lowers the level by 1010 dB, dividing by 100100 lowers it by 2020 dB.

2Try It Yourself

A rock concert measures 110110 dB. Ear plugs reduce the sound intensity reaching your ear to 11000\frac{1}{1000} of its value.

What sound level, in decibels, reaches your ear? (The answer is a whole number, no rounding needed.)

Step 1: Write the mathematical expression

Write the new level as 10log⁡(10111000)10 \log\left(\frac{10^{11}}{1000}\right) and evaluate:

Planning a Savings Goal

Financial planners use logarithms to find how many years an investment needs to reach a goal.

Example:

At 4%4\% per year, money grows by the factor 1.041.04 each year, so the number of years is t=ln⁡(goal/start)ln⁡1.04t = \frac{\ln(\text{goal}/\text{start})}{\ln 1.04}.

3Try It Yourself

You invest 5000 dollars at 4%4\% interest compounded once a year. Your goal is 8000 dollars.

How many full years must you wait until the balance first reaches the goal?

Step 1: Write the mathematical expression

Solve 5000⋅1.04t=80005000 \cdot 1.04^t = 8000, then round up to whole years:

Key Takeaways

  • 1In a logarithm word problem the unknown is an exponent: a time, a magnitude or a level
  • 2pH: pH=−log⁡[H+]\text{pH} = -\log[H^+]; going back: [H+]=10−pH[H^+] = 10^{-\text{pH}}
  • 3Decibels: L=10log⁡(II0)L = 10\log\left(\frac{I}{I_0}\right); every 1010 dB more means 1010 times the intensity
  • 4Richter: magnitudes that differ by dd mean amplitudes in the ratio 10d10^{d}
  • 5Time in growth or decay: isolate the power, take ln⁡\ln of both sides, then t=ln⁡(A/P)ln⁡(1±r)t = \frac{\ln(A/P)}{\ln(1 \pm r)}; doubling time ln⁡2ln⁡(1+r)\frac{\ln 2}{\ln(1+r)}
  • 6Round only at the end, exactly as the problem asks, and check if the situation needs whole years

Frequently Asked Questions

Either one works, as long as you use the same one on top and bottom: log⁡2log⁡1.05=ln⁡2ln⁡1.05≈14.21\frac{\log 2}{\log 1.05} = \frac{\ln 2}{\ln 1.05} \approx 14.21. Formulas for pH and decibels are defined with the common logarithm log⁡\log (base 10), so use log⁡\log there.
Either one works, as long as you use the same one on top and bottom: log⁡2log⁡1.05=ln⁡2ln⁡1.05≈14.21\frac{\log 2}{\log 1.05} = \frac{\ln 2}{\ln 1.05} \approx 14.21. Formulas for pH and decibels are defined with the common logarithm log⁡\log (base 10), so use log⁡\log there.
The equation treats time as continuous, but a bank adds interest only once a year. After 14 years the balance is still below the target, so you need the 15th payment. Always ask whether the situation allows fractions of a year.
Mathematically yes: [H+]=2[H^+] = 2 mol/L gives pH=−log⁡2≈−0.30\text{pH} = -\log 2 \approx -0.30. In everyday chemistry, though, almost all solutions fall between 0 and 14.
The "rule of 72" says doubling time is about 72÷(rate in %)72 \div (\text{rate in }\%). At 5%5\%: 72÷5=14.472 \div 5 = 14.4 years, close to the exact 14.2114.21. Use logarithms when the problem asks for an exact answer.

Glossary

pH
A measure of acidity: pH=−log⁡[H+]\text{pH} = -\log[H^+]. Below 7 is acidic, 7 is neutral, above 7 is basic.
Decibel (dB)
Unit of sound level: L=10log⁡(II0)L = 10\log\left(\frac{I}{I_0}\right), where I0I_0 is the threshold of hearing.
Richter scale
A logarithmic scale for earthquakes: one whole step means 1010 times the amplitude of the ground motion.
Doubling time
The time a growing quantity needs to double: t=ln⁡2ln⁡(1+r)t = \frac{\ln 2}{\ln(1 + r)} for growth by rr per period.
Half-life
The time a decaying quantity needs to fall to half of its value.
Growth factor
The number a quantity is multiplied by each period: 1+r1 + r for growth, 1−r1 - r for decay.

Formula Card

pH

pH=−log⁡[H+]  ⟺  [H+]=10−pH\text{pH} = -\log[H^+] \iff [H^+] = 10^{-\text{pH}}

Acidity from the hydrogen ion concentration

Sound level

L=10log⁡(II0)  ⟺  II0=10L/10L = 10\log\left(\frac{I}{I_0}\right) \iff \frac{I}{I_0} = 10^{L/10}

Decibels from the intensity ratio

Richter comparison

A1A2=10M1−M2\frac{A_1}{A_2} = 10^{M_1 - M_2}

Ratio of amplitudes of two earthquakes

Time to reach a target

t=ln⁡(A/P)ln⁡(1+r)t = \frac{\ln(A/P)}{\ln(1 + r)}

From $A = P(1 + r)^t$; use $1 - r$ for decay

Continuous growth

t=ln⁡(A/P)rt = \frac{\ln(A/P)}{r}

From $A = Pe^{rt}$

Half-life

t=h⋅ln⁡(N/N0)ln⁡0.5t = h \cdot \frac{\ln(N/N_0)}{\ln 0.5}

From $N = N_0 \cdot 0.5^{t/h}$ with half-life $h$

Today's challengeOne problem, three tries

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