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. 3520
B. 24
C. 225
D. 371

Answer :

To solve the problem, we need to find the value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = P e^{rt} \)[/tex] given that [tex]\( f(5) = 288.9 \)[/tex] when [tex]\( r = 0.05 \)[/tex].

Here's how you can work it out step-by-step:

1. Understand the given equation:
[tex]\[ f(t) = P e^{rt} \][/tex]
We know:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]

Substituting these values into the function gives:
[tex]\[ 288.9 = P e^{0.05 \times 5} \][/tex]

2. Calculate the exponent:
[tex]\[ r \times t = 0.05 \times 5 = 0.25 \][/tex]

3. Calculate [tex]\( e^{0.25} \)[/tex]:
The approximate value of [tex]\( e^{0.25} \)[/tex] is about 1.284.

4. Solve for [tex]\( P \)[/tex]:
Rearrange the equation to solve for [tex]\( P \)[/tex]:
[tex]\[ 288.9 = P \times 1.284 \][/tex]

Divide both sides by 1.284:
[tex]\[ P = \frac{288.9}{1.284} \][/tex]

Calculating this gives:
[tex]\[ P \approx 225 \][/tex]

5. Conclusion:
Therefore, the approximate value of [tex]\( P \)[/tex] is 225.

So the correct answer choice is C. 225.