College

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 solve for the approximate value of [tex]\( P \)[/tex] in the function [tex]\( f(t) = P e^{rt} \)[/tex], where [tex]\( f(5) = 288.9 \)[/tex] and [tex]\( r = 0.05 \)[/tex], follow these steps:

1. Understand the Function:
The given equation is [tex]\( f(t) = P e^{rt} \)[/tex]. We're given specific values, [tex]\( f(5) = 288.9 \)[/tex], [tex]\( r = 0.05 \)[/tex], and [tex]\( t = 5 \)[/tex].

2. Substitute Known Values:
We substitute the known values into the equation:
[tex]\[
288.9 = P e^{0.05 \times 5}
\][/tex]

3. Simplify the Exponent:
Calculate the exponent:
[tex]\[
0.05 \times 5 = 0.25
\][/tex]

4. Calculate [tex]\( e^{0.25} \)[/tex]:
Find the value of [tex]\( e^{0.25} \)[/tex], which is approximately [tex]\( 1.284 \)[/tex].

5. Solve for [tex]\( P \)[/tex]:
Substitute this result back into the equation:
[tex]\[
288.9 = P \times 1.284
\][/tex]
To find [tex]\( P \)[/tex], divide both sides by [tex]\( 1.284 \)[/tex]:
[tex]\[
P = \frac{288.9}{1.284}
\][/tex]

6. Calculate [tex]\( P \)[/tex]:
Performing the division gives [tex]\( P \approx 225 \)[/tex].

Therefore, the approximate value of [tex]\( P \)[/tex] is [tex]\( \boxed{225} \)[/tex], which corresponds to option C.