Exponential and Logarithmic Word Problems Practice Test

Model real situations with exponential growth, decay, logarithms, interest, half-life, pH, and measurement scales.

Exponential and Logarithmic Word Problems Practice Test

20 applied growth, decay, finance, science, and logarithm questions.

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Applied Model Mosaic

The hardest part of a word problem is often choosing the model before calculating.

Exponential and logarithmic applications describe repeated multiplicative change, continuous change, half-life, finance, and measurement scales. A strong solution starts by identifying the initial value, growth or decay factor, time unit, and model type. Logarithms enter only when the unknown is trapped in an exponent or when the scale itself is logarithmic.

Repeated growthUse a factor greater than one for equal-percent increases.

Repeated decayUse a factor between zero and one for equal-percent decreases.

InterestChoose periodic compounding or continuous growth.

Logarithmic scaleInterpret orders of magnitude rather than linear differences.

1. Choose the model before substituting numbers

The wording tells you whether change is repeated by a fixed factor, continuous, or organized into repeated half-life periods.

2. Population growth: repeated percentage change bends upward over time

A constant percentage increase is exponential, not linear, because each new increase is applied to a changing quantity.

Illustrative population model

A (t) = 50000 {(1.03)}^{t}

A (10) = 50000 {(1.03)}^{10} \approx 67196

This example starts with fifty thousand and grows by three percent per year. The graph curves upward because the same percentage is applied to a larger base each year.

Population over twenty yearsThe plotted point marks the ten-year value.

The slope increases over time: equal percentage changes produce larger absolute increases as the population grows.

3. Depreciation and medication decay: repeated loss curves downward

A fifteen-percent decrease means eighty-five percent remains after each period.

Illustrative depreciation model

A (t) = 30000 {(0.85)}^{t}

A (5) = 30000 {(0.85)}^{5} \approx 13311

The value drops quickly at first, then more slowly in absolute terms because each later percentage loss is applied to a smaller remaining amount.

Repeated depreciationThe plotted point marks the fifth period.

Exponential decay approaches zero without behaving like a straight-line decrease.

4. Finance: compounding frequency changes the growth path

Periodic compounding and continuous compounding are related but not identical models.

Illustrative investment comparison

A = 5000 {(1 + \frac{0.05}{12})}^{12 t}

A = 5000 e^{0.05 t}

Both models start from the same principal and annual rate. The difference comes from how often interest is applied.

Monthly versus continuous growthTeal = monthly compounding; lilac = continuous compounding.

At the same nominal rate, the continuous model grows slightly faster because accumulation happens without discrete breaks.

5. Half-life: count elapsed half-life periods before evaluating

The phrase “half-life” describes the time interval, not the amount remaining.

Illustrative half-life model

A (t) = 100 {(\frac{1}{2})}^{\frac{t}{6}}

\frac{t}{h}

A (18) = 100 {(\frac{1}{2})}^{3} = 12.5

After eighteen time units with a six-unit half-life, three half-life periods have passed, so the amount has been halved three times.

Half-life decayThe plotted point marks three completed half-life periods.

The decay is multiplicative: each equal half-life interval cuts the current amount in half.

6. Solve for time only after isolating the exponential expression

Logarithms are a solve-back tool when the unknown appears in the exponent.

Four-step solve-backModel first, isolate the exponential factor, take logarithms, then interpret the time value.

1. Target equation

50000 {(1.03)}^{t} = 70000

The model is already written and the target is substituted.

2. Isolate power

{(1.03)}^{t} = \frac{70000}{50000}

Remove the initial value before taking logarithms.

3. Apply logs

t \ln (1.03) = \ln (\frac{70000}{50000})

The exponent becomes an ordinary multiplier.

4. Solve time

t = \frac{\ln (\frac{70000}{50000})}{\ln (1.03)} \approx 11.38

Keep the exact logarithmic ratio until approximation is required.

7. Logarithmic measurement scales turn multiplicative ratios into additive scores

pH, decibels, and earthquake magnitude are not ordinary linear scales. A fixed increase in the reported score represents a multiplicative change in the underlying quantity.

pH

pH = - \log ([H^{+}])

acidicneutralbasic

pH = - \log (10^{- 4}) = 4

The negative logarithm reverses the direction: a smaller hydrogen-ion concentration produces a larger pH value.

Decibels

L = 10 \log (\frac{I}{I_{0}})

referencestrongermuch stronger

10 \log (100) = 20

A hundredfold intensity ratio becomes an additive twenty-unit increase on the decibel scale.

Earthquake magnitude

M = \log (\frac{A}{A_{0}})

smallermoderatelarger

\log (1000) = 3

A thousandfold amplitude ratio corresponds to an increase of three units in this simplified base-ten model.

8. Interpretation matters as much as the calculation

A numerical result is incomplete until it is connected back to the quantities and units in the story.

Linear versus exponential

A_{0} + r t

A_{0} {(1 + r)}^{t}

Repeated percentage change multiplies the current amount; it does not add the same fixed amount each period.

Percent to decimal

3 % = 0.03

A percentage rate must be converted to decimal form before it becomes part of the growth or decay factor.

Doubling language

A (t) = A_{0} 2^{\frac{t}{h}}

When a problem gives a doubling time directly, the exponent counts how many doubling periods have elapsed.

9. Error analysis: most word-problem mistakes begin before the calculator

The wrong model can produce a polished but meaningless numerical answer.

Simple change used instead of repeated multiplication

Constant percentage change is exponential, not linear.

Percent entered as a whole number

Convert the percentage to decimal form before building the factor.

Half-life confused with amount remaining

The half-life is a time interval; the remaining fraction changes after each interval.

Logarithms used before isolation

First isolate the exponential expression, then take logarithms.

Compounding frequency ignored

Monthly, yearly, and continuous models use different structures.

Logarithm ratio reversed

In measurement-scale formulas, preserve the model’s numerator and denominator order.

Final modeling audit

Before accepting a word-problem answer, verify model choice, units, rate conversion, periods, and interpretation.

Did I write a model before substituting numbers?This separates reasoning from arithmetic.

Did I convert percentage change into the correct growth or decay factor?Increase uses one plus the decimal rate; decrease uses one minus it.

Did I count periods or half-life intervals correctly?Match the time unit in the story to the exponent.

Did I choose compound or continuous interest correctly?Use the model that matches the stated compounding method.

Did I isolate the exponential part before applying logarithms?Logarithms solve for exponents only after the power is isolated.

Did I interpret the result with units and context?A model answer should explain what the number means in the original situation.

Exponential and Logarithmic Word Problems Practice Test