Choosing Between Exponential and Linear Depreciation
A student analyzes the resale value of a car that cost $24,000 and, according to a used-car guide, loses 12% of its value each year. She writes V(t) = 24,000(0.12)^t, reports that the car will be worth $2,880 after one year, and concludes that the car 'loses value fast at first and then levels off.' A classmate disagrees and proposes L(t) = 24,000 - 2,880t, arguing that 'it loses the same $2,880 every year, so a line is simpler and just as good.' A table of values for the model year reads $24,000; $21,120; $18,586; $16,355. The student's one-year figure of $2,880 happens to equal the actual first-year loss, which is why her model looks right to her. The tutor must decide which model the table supports, what the base 0.12 actually represents, and how to answer the owner's question, 'When will it be worth half of what I paid?'
1. Diagnose the student's model, write the correct exponential model, and explain the meaning of both 0.12 and 0.88 in this context.
Read the worked answer
The student used the decay rate as the multiplier. In V(t) = a·b^t, b is the factor by which the value is multiplied each year. Losing 12% means keeping 88%, so b = 1 - 0.12 = 0.88 and the correct model is V(t) = 24,000(0.88)^t.
- 0.12 is the fraction of value lost each year (the decay rate).
- 0.88 is the fraction of value retained each year (the decay factor).
Her model gives 24,000(0.12) = $2,880 after one year, which would mean the car kept only 12% of its value - an 88% loss. That $2,880 is actually the amount lost in year one: 24,000 - 21,120 = 2,880. She calculated the loss and reported it as the value. The correct year-one value is 24,000(0.88) = $21,120, matching the table.
2. Evaluate the classmate's linear model against the table and explain which model the data support and why, including a limitation of each.
Read the worked answer
The table shows constant ratios, not constant differences: 21,120/24,000 = 0.88, 18,586/21,120 ≈ 0.88, 16,355/18,586 ≈ 0.88, while the differences are 2,880, 2,534, 2,231 - shrinking each year. Constant ratio is the signature of exponential decay; constant difference is the signature of a line. The data therefore support the exponential model.
The linear model also fails on its own terms: it predicts a value of 24,000 - 2,880(5) = $9,600 at five years (versus about $12,665 from 0.88^5), and it reaches zero at t = 2,880/24,000... i.e., t = 24,000/2,880 ≈ 8.3 years and then predicts negative values, which is impossible for a resale price.
Limitations: the exponential model never reaches zero and will overstate value for a very old car, and real prices depend on mileage and the market, so both models are approximations outside the tabulated years. A linear model could be locally reasonable over a short window but not over the car's life.
3. Answer the owner's question - when the car is worth half its purchase price - showing the reasoning without relying only on a calculator guess.
Read the worked answer
Half of $24,000 is $12,000, so solve 24,000(0.88)^t = 12,000, i.e., 0.88^t = 0.5.
Bracket with a table: 0.88^5 ≈ 0.5277 (value ≈ $12,665) and 0.88^6 ≈ 0.4644 (value ≈ $11,146). The half-value is crossed between 5 and 6 years.
Solve exactly with logarithms: t = ln(0.5)/ln(0.88) ≈ (-0.6931)/(-0.1278) ≈ 5.4 years. Check: 0.88^5.4 ≈ 0.4996, so the value is about $11,990 - half the purchase price.
Interpretation for the owner: the car is worth about half its original price roughly 5.4 years after purchase, under the assumption that the 12% annual rate continues. Note that the answer is not 12% of some fixed number of years and not the linear model's answer (which would place the half-value at t = 12,000/2,880 ≈ 4.2 years), because the dollar loss shrinks each year in the exponential model.