Sure, let's solve this step-by-step.
### a) Finding the Exponential Growth Rate [tex]\( k \)[/tex]
Given:
- Initial value in 1980, [tex]\( V_0 = 212 \)[/tex]
- Value in 1990, [tex]\( V_{10} = 418 \)[/tex]
- Time period, [tex]\( t = 10 \)[/tex] years
We are assuming that the growth is exponential. The formula for exponential growth is:
[tex]\[ V(t) = V_0 e^{kt} \][/tex]
In this problem, we want to find the growth rate [tex]\( k \)[/tex].
First, we use the values given to form the equation:
[tex]\[ 418 = 212 e^{10k} \][/tex]
To solve for [tex]\( k \)[/tex], we need to isolate [tex]\( k \)[/tex]:
1. Divide both sides of the equation by 212:
[tex]\[ \frac{418}{212} = e^{10k} \][/tex]
2. Simplify the fraction on the left:
[tex]\[ 1.9717 \approx e^{10k} \][/tex]
3. To get rid of the exponential, take the natural logarithm (ln) of both sides:
[tex]\[ \ln (1.9717) = \ln (e^{10k}) \][/tex]
4. Apply the property of logarithms [tex]\(\ln (e^x) = x\)[/tex]:
[tex]\[ \ln (1.9717) = 10k \][/tex]
5. Solve for [tex]\( k \)[/tex] by dividing both sides by 10:
[tex]\[ k = \frac{\ln (1.9717)}{10} \][/tex]
Using a calculator to find the natural logarithm:
[tex]\[ k \approx \frac{0.6789}{10} = 0.0679 \][/tex]
Rounded to the nearest thousandth:
[tex]\[ k \approx 0.068 \][/tex]
### b) Finding the Exponential Growth Function [tex]\( V(t) \)[/tex]
Now that we have the growth rate [tex]\( k \approx 0.068 \)[/tex], let's find the exponential growth function:
The general form of the exponential growth function is:
[tex]\[ V(t) = V_0 e^{kt} \][/tex]
Plugging in the known values:
- [tex]\( V_0 = 212 \)[/tex]
- [tex]\( k = 0.068 \)[/tex]
Thus, the exponential growth function is:
[tex]\[ V(t) = 212 e^{0.068t} \][/tex]
Therefore, the solutions are:
1. The exponential growth rate [tex]\( k \approx 0.068 \)[/tex]
2. The exponential growth function:
[tex]\[ V(t) = 212 e^{0.068t} \][/tex]
