Question

A car was valued at $31,000 in the year 1995. The value depreciated to $13,000 by the year 2007.

A) What was the annual rate of change between 1995 and 2007?
r=

Answer 1

To find the annual rate of change (depreciation rate) between 1995 and 2007, we can use the formula for the annual rate of change in the context of exponential decay:

\[
V(t) = V_0 \cdot e^{-rt}
\]

where:
- \( V(t) \) is the value of the car at time \( t \),
- \( V_0 \) is the initial value (1995),
- \( r \) is the annual rate of change (depreciation rate), and
- \( t \) is the number of years since the initial value was determined.

1. **Identify the values:**
- Initial value, \( V_0 = 31000 \)
- Value in 2007, \( V(12) = 13000 \) (since 2007 is 12 years after 1995)
- \( t = 12 \)

2. **Set up the equation:**
\[
13000 = 31000 \cdot e^{-12r}
\]

3. **Solve for \( e^{-12r} \)**:
\[
\frac{13000}{31000} = e^{-12r}
\]
\[
\frac{13}{31} = e^{-12r}
\]

4. **Take the natural logarithm of both sides:**
\[
\ln\left(\frac{13}{31}\right) = -12r
\]

5. **Now, solve for \( r \):**
\[
r = -\frac{1}{12} \ln\left(\frac{13}{31}\right)
\]

6. **Calculate the logarithmic value:**
- Calculate \( \frac{13}{31} \):
\[
\frac{13}{31} \approx 0.4194
\]
- Take the natural logarithm:
\[
\ln(0.4194) \approx -0.872
\]
- Now substitute this back into the equation for \( r \):
\[
r = -\frac{1}{12} \cdot (-0.872) \approx \frac{0.872}{12} \approx 0.07267
\]

7. **Convert \( r \) to a percentage:**
\[
r \approx 0.07267 \times 100 \approx 7.27\%
\]

Thus, the annual rate of change (depreciation rate) is approximately **7.27%**.