A doll sold for $249 in 1977 and was sold again in 1986 for $472. Assume that the growth in the value V of the collector's item was exponential. H a) Find the value k of the exponential growth rate. Assume V = 249. k=0 (Round to the nearest thousandth.)
A doll sold for $249 in 1977 and was sold again in 1986 for $472. Assume that the growth in the value V of the collector's item was exponential. H a) Find the value k of the exponential growth rate. Assume V = 249. k=0 (Round to the nearest thousandth.)
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Author:James Stewart
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![**Exponential Growth Rate Calculation**
A doll was sold for $249 in 1977 and was sold again in 1986 for $472. Assume that the growth in the value \( V \) of the collector's item was exponential.
**Question:**
a) Find the value \( k \) of the exponential growth rate. Assume \( V_0 = 249 \).
\[ k = \_\_\_\_\_\_ \]
(Round to the nearest thousandth.)
---
### Explanation
To find the exponential growth rate \( k \), we must use the exponential growth formula:
\[ V(t) = V_0 e^{kt} \]
Where:
- \( V(t) \) is the final value after time \( t \).
- \( V_0 \) is the initial value.
- \( k \) is the exponential growth rate.
- \( t \) is the time period over which the growth occurs.
- \( e \) is the base of the natural logarithm.
Given:
- Initial value, \( V_0 = 249 \).
- Final value after time, \( V(t) = 472 \) (in 1986).
- Time period, \( t = 1986 - 1977 = 9 \) years.
We need to solve for \( k \).
### Calculation Steps:
1. Substitute the known values into the formula:
\[ 472 = 249 e^{9k} \]
2. Divide both sides of the equation by 249:
\[ \frac{472}{249} = e^{9k} \]
3. Simplify the fraction:
\[ 1.89558 = e^{9k} \]
4. Apply the natural logarithm (ln) to both sides to eliminate the exponential:
\[ \ln(1.89558) = 9k \]
5. Solve for \( k \):
\[ k = \frac{\ln(1.89558)}{9} \]
\[ k \approx 0.074 \]
(rounded to the nearest thousandth).
So, the exponential growth rate \( k \) is approximately **0.074**.
---
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---
*Note: The question involves finding the exponential growth rate given the initial and final values over a given time period. Ensure to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3230e3d-dd38-497e-8c1e-7c1c5570828c%2F560fad5d-34f3-426b-a6cb-1ba5f494b0a2%2Fk8cidz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exponential Growth Rate Calculation**
A doll was sold for $249 in 1977 and was sold again in 1986 for $472. Assume that the growth in the value \( V \) of the collector's item was exponential.
**Question:**
a) Find the value \( k \) of the exponential growth rate. Assume \( V_0 = 249 \).
\[ k = \_\_\_\_\_\_ \]
(Round to the nearest thousandth.)
---
### Explanation
To find the exponential growth rate \( k \), we must use the exponential growth formula:
\[ V(t) = V_0 e^{kt} \]
Where:
- \( V(t) \) is the final value after time \( t \).
- \( V_0 \) is the initial value.
- \( k \) is the exponential growth rate.
- \( t \) is the time period over which the growth occurs.
- \( e \) is the base of the natural logarithm.
Given:
- Initial value, \( V_0 = 249 \).
- Final value after time, \( V(t) = 472 \) (in 1986).
- Time period, \( t = 1986 - 1977 = 9 \) years.
We need to solve for \( k \).
### Calculation Steps:
1. Substitute the known values into the formula:
\[ 472 = 249 e^{9k} \]
2. Divide both sides of the equation by 249:
\[ \frac{472}{249} = e^{9k} \]
3. Simplify the fraction:
\[ 1.89558 = e^{9k} \]
4. Apply the natural logarithm (ln) to both sides to eliminate the exponential:
\[ \ln(1.89558) = 9k \]
5. Solve for \( k \):
\[ k = \frac{\ln(1.89558)}{9} \]
\[ k \approx 0.074 \]
(rounded to the nearest thousandth).
So, the exponential growth rate \( k \) is approximately **0.074**.
---
For additional help, you can click on the following options:
- **Help me solve this**
- **View an example**
- **Get more help**
---
*Note: The question involves finding the exponential growth rate given the initial and final values over a given time period. Ensure to
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