Respuesta :
We have the function
[tex]p(t)=550(1-e^{-0.039t})[/tex]Therefore we want to determine when we have
[tex]p(t_0)=550[/tex]It means that the term
[tex]e^{-0.039t}[/tex]Must go to zero, then let's forget the rest of the function for a sec and focus only on this term
[tex]e^{-0.039t}\rightarrow0[/tex]But for which value of t? When we have a decreasing exponential, it's interesting to input values that are multiples of the exponential coefficient, if we have 0.039 in the exponential, let's define that
[tex]\alpha=\frac{1}{0.039}[/tex]The inverse of the number, but why do that? look what happens when we do t = α
[tex]e^{-0.039t}\Rightarrow e^{-0.039\alpha}\Rightarrow e^{-1}=\frac{1}{e}[/tex]And when t = 2α
[tex]e^{-0.039t}\Rightarrow e^{-0.039\cdot2\alpha}\Rightarrow e^{-2}=\frac{1}{e^2}[/tex]We can write it in terms of e only.
And we can find for which value of α we have a small value that satisfies
[tex]e^{-0.039t}\approx0[/tex]Only using powers of e
Let's write some inverse powers of e:
[tex]\begin{gathered} \frac{1}{e}=0.368 \\ \\ \frac{1}{e^2}=0.135 \\ \\ \frac{1}{e^3}=0.05 \\ \\ \frac{1}{e^4}=0.02 \\ \\ \frac{1}{e^5}=0.006 \end{gathered}[/tex]See that at t = 5α we have a small value already, then if we input p(5α) we can get
[tex]\begin{gathered} p(5\alpha)=550(1-e^{-0.039\cdot5\alpha}) \\ \\ p(5\alpha)=550(1-0.006) \\ \\ p(5\alpha)=550(1-0.006) \\ \\ p(5\alpha)=550\cdot0.994 \\ \\ p(5\alpha)\approx547 \end{gathered}[/tex]That's already very close to 550, if we want a better approximation we can use t = 8α, which will result in 549.81, which is basically 550.
Therefore, we can use t = 5α and say that 3 people are not important for our case, and say that it's basically 550, or use t = 8α and get a very close value.
In both cases, the decimal answers would be
[tex]\begin{gathered} 5\alpha=\frac{5}{0.039}=128.2\text{ minutes (good approx)} \\ \\ 8\alpha=\frac{8}{0.039}=205.13\text{ minutes (even better approx)} \end{gathered}[/tex]