High School

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. 225
B. 24
C. 3520
D. 371

Answer :

To solve this problem, we need to find the initial value [tex]\( P \)[/tex] given the function [tex]\( f(t) = P e^{rt} \)[/tex]. We're told that [tex]\( f(5) = 288.9 \)[/tex] when [tex]\( r = 0.05 \)[/tex]. Let's determine [tex]\( P \)[/tex] step-by-step:

1. Understand the function and what is given:

The function is [tex]\( f(t) = P e^{rt} \)[/tex].

2. Substitute the known values into the function:

We're given that [tex]\( f(5) = 288.9 \)[/tex], [tex]\( r = 0.05 \)[/tex], and [tex]\( t = 5 \)[/tex]. So we set up 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]

Therefore, the equation becomes:

[tex]\[
288.9 = P e^{0.25}
\][/tex]

4. Solve for [tex]\( P \)[/tex]:

We need to find [tex]\( e^{0.25} \)[/tex]. This value is approximately [tex]\( 1.284 \)[/tex].

5. Calculate [tex]\( P \)[/tex] by isolating it:

Divide both sides of the equation by [tex]\( e^{0.25} \)[/tex]:

[tex]\[
P = \frac{288.9}{e^{0.25}}
\][/tex]

[tex]\[
P \approx \frac{288.9}{1.284} \approx 225
\][/tex]

Thus, the approximate value of [tex]\( P \)[/tex] is 225, which corresponds to option A.