Question

Your family purchases a new car for \(\$ 20,000\). Its value decreases by \(15 \%\) each year. During what interval does the car's value exceed \(\$ 10,000\) ?

Short answer

The car's value will exceed $10,000 for the interval from the time of purchase until approximately 4 years later.

Step-by-step solution

Understand the concept of depreciation and Identify the values

In this problem, the original value of the car is $20,000 and it depreciates by 15% each year. The aim is to determine when the car's value drops below $10,000.

Set up the depreciation formula

The standard formula to calculate depreciation, specifically for exponential depreciation, is: \( P = P_0 (1 - r)^t \)where \( P \) is the future value of the investment,\( P_0 \) is the initial amount of the investment,\( r \) is the depreciation rate (in decimal form),\( t \) is the time in years.Here, \( P_0 = $20,000 \), \( r = 15% = 0.15 \) and \( P = $10,000 \)

Substitute the values into the formula

Substituting these values into the formula, we get: $10,000 = $20,000(1 - 0.15)^t

Solve the equation

To solve for \( t \), first divide both sides by $20,000. This gives: \[ 0.5 = (0.85)^t \] Now, apply logarithm on both sides: \( log(0.5) = t * log(0.85) \). Solving for \( t \), we get: \( t = \frac{log(0.5)}{log(0.85)} \)

Calculate the time

After solving this, approximately \( t = 4.57 \) years . This means the car’s value will exceed $10,000 for about 4 years, as we can only count full years

Interpret the result

Therefore, the car's value will exceed $10,000 for the interval from the time of purchase until 4 years later.

Key concepts

Depreciation Formula

When you make a purchase like a car, it loses value over time, which is known as depreciation. This is an important concept in finance, especially when planning to sell or trade in assets later.

The depreciation formula helps to calculate how much value an investment will have after a certain period. It is especially useful for assets that lose value at a constant percentage rate annually. The formula for calculating depreciation is:

• \( P = P_0 (1 - r)^t \)

Where:

• \( P \) = the future value of the asset
• \( P_0 \) = the original purchase price
• \( r \) = the annual depreciation rate (as a decimal)
• \( t \) = the number of years

By inserting the known values into this formula, it becomes possible to predict the value of the car (or any other asset) after a certain number of years.

Understanding this formula is crucial as it shows that even though an asset is initially purchased at a given price, its worth changes over time due to depreciation.

Exponential Decay

Exponential decay is a decrease in value at a consistent rate, such as a car losing 15% of its value each year. This concept shows how quickly an asset can lose value in the initial years compared to later years. With exponential decay, the rate of depreciation is proportional to the current value. So, as the value decreases each year, the absolute amount the asset loses also decreases, even if the rate stays the same.

For instance, in the car depreciation example, the car's value decreases from \(\\(20,000\) yearly by 15%. However, each year, the decrement is smaller than the previous one, because 15% of the diminished value is less in absolute terms than 15% of the original value.

• Year 1: \(15\%\) of \(\\)20,000 = \\(3,000\)
• Year 2: \(15\%\) of \(\\)17,000 = \$2,550\)

Understanding exponential decay is key when evaluating investments and predicting future values accurately.

Logarithmic Calculation

Logarithmic calculation comes into play when solving exponential equations, especially when determining the time it takes for an exponential process to reach a certain value. In the context of the car depreciation problem, logarithms help us find out when the car's value will fall below \(\\(10,000\).

After setting up the formula:

• \( 0.5 = (0.85)^t \)

We need to isolate \(t\) to find how many years will pass before reaching the desired value. This is done by taking the logarithm of both sides of the equation:

• \( \log(0.5) = t \times \log(0.85) \)

To solve for \(t\), divide both sides by \(\log(0.85)\):

• \( t = \frac{\log(0.5)}{\log(0.85)} \)

This equation ensures you correctly calculate the number of years required for the car's value to reduce below a specific threshold, such as \(\\)10,000\).

Familiarity with logarithms is important as it allows you to navigate these calculations confidently and determine timeframes linked to exponential growth and decay.