High School

If [tex]$f(5)=288.9$[/tex] when [tex]$r=0.05$[/tex] for the function [tex]$f(t)=P e^{rt}$[/tex], then what is the approximate value of [tex]$P$[/tex]?

A. 24
B. 371
C. 225
D. 3520

Answer :

To find the approximate value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = Pe^{rt} \)[/tex] given that [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex], follow these steps:

1. Identify the Given Information:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]

2. Write the Formula for Exponential Growth:
The function is expressed as:
[tex]\[
f(t) = Pe^{rt}
\][/tex]

3. Substitute the Values into the Formula:
Plug the given values into the equation:
[tex]\[
288.9 = Pe^{0.05 \times 5}
\][/tex]

4. Calculate [tex]\( e^{rt} \)[/tex]:
First, compute the exponent:
[tex]\[
e^{0.05 \times 5} = e^{0.25}
\][/tex]

5. Solve for [tex]\( P \)[/tex]:
Rearrange the formula to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]

6. Calculate the Approximate Value of [tex]\( P \)[/tex]:
Calculate the value using the approximation for [tex]\( e^{0.25} \)[/tex], and subsequently find:
[tex]\[
P \approx \frac{288.9}{1.284} \approx 225
\][/tex]

The approximate value of [tex]\( P \)[/tex] is 225. Therefore, the correct answer is option C.