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If [tex]f(3) = 191.5[/tex] when [tex]r = 0.03[/tex] for the function [tex]f(t) = P e^t[/tex], then what is the approximate value of [tex]P[/tex]?

A. 175
B. 210
C. 471
D. 78

Answer :

We are given the function

[tex]$$
f(t) = P e^{rt},
$$[/tex]

with [tex]\( r = 0.03 \)[/tex] and [tex]\( f(3) = 191.5 \)[/tex]. Substituting [tex]\( t = 3 \)[/tex] into the function, we have

[tex]$$
191.5 = P e^{0.03 \times 3}.
$$[/tex]

Simplify the exponent:

[tex]$$
0.03 \times 3 = 0.09.
$$[/tex]

So, the equation becomes

[tex]$$
191.5 = P e^{0.09}.
$$[/tex]

To solve for [tex]\( P \)[/tex], divide both sides of the equation by [tex]\( e^{0.09} \)[/tex]:

[tex]$$
P = \frac{191.5}{e^{0.09}}.
$$[/tex]

Evaluating [tex]\( e^{0.09} \)[/tex] gives approximately [tex]\( 1.09417 \)[/tex], so

[tex]$$
P \approx \frac{191.5}{1.09417} \approx 175.
$$[/tex]

Thus, the approximate value of [tex]\( P \)[/tex] is [tex]\( 175 \)[/tex].

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