Question

The value \(V\) of a Chevy Cruze LT that is \(t\) years old can be modeled by \(V(t)=19,200(0.82)^{t}\) (a) According to the model, when will the car be worth 12,000 dollar ? (b) According to the model, when will the car be worth 9000 dollar ? (c) According to the model, when will the car be worth 3000 dollar ?

Short answer

(a) Approximately 2.76 years. (b) Approximately 4.23 years. (c) Approximately 8.35 years.

Step-by-step solution

Set up the equation

For each part, set up the equation by substituting the given value into the equation \(V(t) = 19200(0.82)^t\).

(a) Set up the equation for \(12,000

Substitute 12000 into the equation: \(12000 = 19200(0.82)^t\).

(a) Solve for \)t\(

Divide both sides by 19200: \(\frac{12000}{19200} = (0.82)^t\). Simplify to get \(0.625 = (0.82)^t\). Take the natural logarithm of both sides: \(\ln(0.625) = t \cdot \ln(0.82)\). Solve for \(t\) by dividing: \(t = \frac{\ln(0.625)}{\ln(0.82)} \approx 2.76\).

(b) Set up the equation for \)9000

Substitute 9000 into the equation: \(9000 = 19200(0.82)^t\).

(b) Solve for \(t\)

Divide both sides by 19200: \(\frac{9000}{19200} = (0.82)^t\). Simplify to get \(0.46875 = (0.82)^t\). Take the natural logarithm of both sides: \(\ln(0.46875) = t \cdot \ln(0.82)\). Solve for \(t\) by dividing: \(t = \frac{\ln(0.46875)}{\ln(0.82)} \approx 4.23\).

(c) Set up the equation for \(3000

Substitute 3000 into the equation: \(3000 = 19200(0.82)^t\).

(c) Solve for \)t$

Divide both sides by 19200: \(\frac{3000}{19200} = (0.82)^t\). Simplify to get \(0.15625 = (0.82)^t\). Take the natural logarithm of both sides: \(\ln(0.15625) = t \cdot \ln(0.82)\). Solve for \(t\) by dividing: \(t = \frac{\ln(0.15625)}{\ln(0.82)} \approx 8.35\).

Key concepts

Exponential Decay

To understand how exponential decay works, let’s look at the given problem about a car’s value decreasing over time. Exponential decay models are commonly used to represent how something decreases at a consistent percentage rate. In this example, the value of a Chevy Cruze LT depreciates by 18% (since 100% - 82% = 18%) each year. The general formula for exponential decay is:
\[ V(t) = V_0 \times (b)^t \]
Here, \(V_0\) is the initial value (19,200 in this case), \(b\) is the decay factor (0.82), and \(t\) is time in years.
Exponential decay can be visualized as follows: with each passing year, the value of the car is multiplied by a factor less than 1. This results in an ever-decreasing value over time.

Natural Logarithm

Natural logarithms (often denoted as ln) come into play when we need to solve equations involving exponential decay. They help us 'undo' the exponential function so we can isolate and solve for the variable exponent. In mathematical terms, if we have an equation of the form:
\[ a = b^t \]
Taking the natural logarithm of both sides lets us bring down the exponent:
\[ \text{ln}(a) = t \times \text{ln}(b) \]
This property of logarithms is incredibly useful because it allows us to easily solve for the exponent \(t\). For instance, to solve for \(t\) when the car’s value is \(12,000, \)9000, or \(3000, we use:
\[ t = \frac{\text{ln}(0.625)}{\text{ln}(0.82)} \]
for \)12,000, and similar setups for the other values.

Solving Equations

Let’s break down solving the equation step-by-step. First, substitute the given value into the equation. For example, to find when the car is worth \(12,000, we plug 12,000 into our model:
\[ 12000 = 19200(0.82)^t \]
Next, divide both sides by 19200 to isolate the decay factor:
\[ \frac{12000}{19200} = 0.82^t \]
Simplify the left side:
\[ 0.625 = 0.82^t \]
Now, take the natural logarithm of both sides to solve for \)t\(:
\[ \text{ln}(0.625) = t \times \text{ln}(0.82) \]
Finally, divide by \)\text{ln}(0.82)\( to isolate \)t\(:
\[ t = \frac{\text{ln}(0.625)}{\text{ln}(0.82)} \]
Simplifying this using a calculator gives approximately 2.76 years. Repeat similar steps for \)9000 and $3000 to find the time when the car's value reaches those amounts. By following these steps, solving exponential decay problems becomes more manageable!