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

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


Sagot :

To find the value of [tex]\( P \)[/tex] given the function [tex]\( f(t) = P e^{r t} \)[/tex], we can follow these steps:

1. Identifying Known Values:
- [tex]\( f(5) = 288.9 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( t = 5 \)[/tex]

2. Substituting Known Values into the Function:
The function is [tex]\( f(t) = P e^{r t} \)[/tex]. With [tex]\( t = 5 \)[/tex], the equation becomes:
[tex]\[ f(5) = P e^{0.05 \cdot 5} \][/tex]

3. Calculating the Exponential Term:
The exponent is:
[tex]\[ 0.05 \times 5 = 0.25 \][/tex]
So, [tex]\( e^{0.25} \)[/tex].

4. Finding [tex]\( e^{0.25} \)[/tex]:
The value of [tex]\( e^{0.25} \)[/tex] is approximately [tex]\( 1.2840254166877414 \)[/tex].

5. Rewriting the Equation:
Now, we have:
[tex]\[ 288.9 = P \times 1.2840254166877414 \][/tex]

6. Solving for [tex]\( P \)[/tex]:
To isolate [tex]\( P \)[/tex], divide both sides of the equation by [tex]\( 1.2840254166877414 \)[/tex]:
[tex]\[ P = \frac{288.9}{1.2840254166877414} \approx 224.99554622932885 \][/tex]

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

Answer:
C. 225