Let's do an example together that is an exponential decay model.

Problem

The cost of a new car is [tex]\$ 32,000[/tex]. It depreciates at a rate of [tex]15 \%[/tex] per year. This means that it loses [tex]15 \%[/tex] of its value each year.

1. Draw a graph of the car's value against time in years.
2. Find the formula that gives the value of the car in terms of time.
3. Find the value of the car when it is four years old.

Solution

This is an exponential decay function. Start by making a table of values. To fill in the values, we start with [tex]\$32,000[/tex] when [tex]t=0[/tex]. Then we multiply the value of the car by [tex]85 \%[/tex] for each passing year. (Since the car loses [tex]15 \%[/tex] of its value, it keeps [tex]85 \%[/tex] of its value. [tex]100 - 15 = 85[/tex].) Remember [tex]85 \% = 0.85[/tex].

\begin{tabular}{|l|l|}
\hline
Time (years) & Value (Thousands) \\
\hline
0 & 32 \\
\hline
1 & 27.2 \\
\hline
2 & 23.1 \\
\hline
3 & 19.7 \\
\hline
4 & 16.7 \\
\hline
5 & 14.2 \\
\hline
\end{tabular}

The general formula is [tex]y = a(b)^x[/tex]. In this case, [tex]y[/tex] is the value of the car, [tex]x[/tex] is the time in years, [tex]a = 32,000[/tex] (the starting amount in thousands), and [tex]b = 0.85[/tex] (since we multiply the value in any year by this factor to get the value of the car in the following year). The formula for this problem is:
[tex]
y = 32000(0.85)^x
[/tex]

Finally, to find the value of the car when it is four years old, we use [tex]x = 4[/tex] in the formula. Remember the value is in thousands.
[tex]
y = 32000(0.85)^4 = 16,731
[/tex]

At 4 years old, we expect the car to be worth [tex]\$ 16,731[/tex].



Answer :

Sure, let's go through the solution step-by-step.

First, let's make the table of values showing how the car's value changes each year:

| Time (Years) | Value (\[tex]$) | |--------------|-------------------| | 0 | 32,000 | | 1 | 27,200 | | 2 | 23,120 | | 3 | 19,652 | | 4 | 16,704.2 | | 5 | 14,198.6 | Explanation of the table: - At year 0, the value of the car is \$[/tex]32,000.
- At year 1, the value of the car is [tex]\( 32,000 \times 0.85 = 27,200 \)[/tex].
- At year 2, the value of the car is [tex]\( 27,200 \times 0.85 = 23,120 \)[/tex].
- At year 3, the value of the car is [tex]\( 23,120 \times 0.85 = 19,652 \)[/tex].
- At year 4, the value of the car is [tex]\( 19,652 \times 0.85 = 16,704.2 \)[/tex].
- At year 5, the value of the car is [tex]\( 16,704.2 \times 0.85 = 14,198.6 \)[/tex].

Finding the formula:

The general exponential decay formula is given by [tex]\( y = a \cdot b^x \)[/tex], where:
- [tex]\( y \)[/tex] is the value of the car.
- [tex]\( a \)[/tex] is the initial value of the car, which is \[tex]$32,000. - \( b \) is the rate at which the value is retained each year, which is 0.85 (since it retains 85% of its value each year). - \( x \) is the time in years. So, the formula for the car's value in terms of time is: \[ y = 32,000 \cdot (0.85)^x \] Finding the value of the car when it is four years old: To find the value of the car when it is four years old, we substitute \( x = 4 \) into the formula: \[ y = 32,000 \cdot (0.85)^4 \] Calculating the value, we get: \[ y = 32,000 \cdot (0.85)^4 = 32,000 \cdot 0.52200625 \approx 16,704.2 \] Thus, the value of the car when it is 4 years old is approximately: \[ \$[/tex]16,704.2 \]

So, the final answer is that at 4 years old, we expect the car to be worth [tex]$16,704.2$[/tex].

The step-by-step detailed solution includes:
1. Creating a table with value calculations over the years.
2. Deriving the general exponential decay formula.
3. Using the formula to find the specific value at 4 years old.

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