Answer :
- Substitute $t=5$ into the function $f(t) = Pe^t$, resulting in $f(5) = Pe^5$.
- Use the given value $f(5) = 288.9$ to get $288.9 = Pe^5$.
- Solve for $P$ by dividing both sides by $e^5$, so $P = \frac{288.9}{e^5}$.
- Calculate $P \approx \frac{288.9}{1.284} \approx 225$, assuming the function is $f(t) = Pe^{rt}$ where $r=0.05$.
- The approximate value of $P$ is $\boxed{225}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t) = P e^t$ and the information that $f(5) = 288.9$. Our goal is to find the approximate value of $P$.
2. Substituting the Value of t
We substitute $t=5$ into the function, which gives us $f(5) = P e^5$. Since we know that $f(5) = 288.9$, we can write the equation $288.9 = P e^5$.
3. Isolating P
To solve for $P$, we need to divide both sides of the equation by $e^5$. This gives us $P = \frac{288.9}{e^5}$.
4. Calculating e^5
Now, we need to calculate the value of $e^5$. The result of this calculation is approximately $148.413$.
5. Calculating P
Finally, we divide $288.9$ by $148.413$ to find the value of $P$. $P = \frac{288.9}{148.413} \approx 1.947$. Among the given options, the closest value to 1.947 is not present. However, let's re-evaluate the problem statement. The problem states $f(5) = 288.9$ when $r = 0.05$. It seems $r$ is irrelevant here. Let's proceed with the calculation.
6. Considering the rate r
Since none of the options are close to 1.947, there might be a typo in the problem. Let's assume the function is $f(t) = P e^{rt}$ where $r=0.05$. Then $f(5) = P e^{0.05 Imes 5} = P e^{0.25} = 288.9$. Therefore, $P = \frac{288.9}{e^{0.25}}$. Let's calculate $e^{0.25}$.
7. Calculating P with r
The value of $e^{0.25}$ is approximately $1.284$. Then $P = \frac{288.9}{1.284} \approx 225$.
### Examples
Exponential functions are used to model population growth, radioactive decay, and compound interest. In finance, understanding exponential growth helps in calculating investment returns over time. For instance, if you invest a certain amount of money at a fixed interest rate, the exponential function can predict how your investment will grow over the years. This concept is crucial for making informed financial decisions.
- Use the given value $f(5) = 288.9$ to get $288.9 = Pe^5$.
- Solve for $P$ by dividing both sides by $e^5$, so $P = \frac{288.9}{e^5}$.
- Calculate $P \approx \frac{288.9}{1.284} \approx 225$, assuming the function is $f(t) = Pe^{rt}$ where $r=0.05$.
- The approximate value of $P$ is $\boxed{225}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t) = P e^t$ and the information that $f(5) = 288.9$. Our goal is to find the approximate value of $P$.
2. Substituting the Value of t
We substitute $t=5$ into the function, which gives us $f(5) = P e^5$. Since we know that $f(5) = 288.9$, we can write the equation $288.9 = P e^5$.
3. Isolating P
To solve for $P$, we need to divide both sides of the equation by $e^5$. This gives us $P = \frac{288.9}{e^5}$.
4. Calculating e^5
Now, we need to calculate the value of $e^5$. The result of this calculation is approximately $148.413$.
5. Calculating P
Finally, we divide $288.9$ by $148.413$ to find the value of $P$. $P = \frac{288.9}{148.413} \approx 1.947$. Among the given options, the closest value to 1.947 is not present. However, let's re-evaluate the problem statement. The problem states $f(5) = 288.9$ when $r = 0.05$. It seems $r$ is irrelevant here. Let's proceed with the calculation.
6. Considering the rate r
Since none of the options are close to 1.947, there might be a typo in the problem. Let's assume the function is $f(t) = P e^{rt}$ where $r=0.05$. Then $f(5) = P e^{0.05 Imes 5} = P e^{0.25} = 288.9$. Therefore, $P = \frac{288.9}{e^{0.25}}$. Let's calculate $e^{0.25}$.
7. Calculating P with r
The value of $e^{0.25}$ is approximately $1.284$. Then $P = \frac{288.9}{1.284} \approx 225$.
### Examples
Exponential functions are used to model population growth, radioactive decay, and compound interest. In finance, understanding exponential growth helps in calculating investment returns over time. For instance, if you invest a certain amount of money at a fixed interest rate, the exponential function can predict how your investment will grow over the years. This concept is crucial for making informed financial decisions.
