Respuesta :
Answer:
So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).
Step-by-step explanation:
Given:
A bacteria culture with constant relative rate.
No. of bacteria = [tex]400[/tex] Time ([tex]t[/tex]) = [tex]2[/tex] hrs
No.of bacteria = [tex]50,000[/tex] Time ([tex]t[/tex]) = 8 hrs
Here the growth is exponential.
Formula to be used.
- [tex]P(t)=P_0e^k^t[/tex] where [tex]P(t)[/tex] = Number of bacteria after time ([tex]t[/tex]).
[tex]P_0[/tex] = Initial population
According to the question.
[tex]P(2)=P_0e^2^k=400[/tex] ...equation (i)
[tex]P(8)=P_0e^8^k =50,000[/tex] ...equation (ii)
Dividing equation (ii) with (i)
⇒ [tex]\frac{P_0e^8^k}{P_0e^2^k} =\frac{50,000}{400}[/tex]
⇒ [tex]e^6^k=125[/tex]
⇒ [tex](e^k)^6=125[/tex]
Now using [tex]\ln[/tex] function both sides.
⇒ [tex]\ln (e^k)^6=\ln(125)[/tex]
⇒ [tex]6\ln(e^k)=\ln (125)[/tex] using [tex]\ln x^a=a\ln x[/tex] rule
⇒ [tex]\ln e^k=\frac{\ln 125}{6}[/tex]
And from logarithmic rule [tex]\ln(e) = 1[/tex]
⇒ [tex]k=\frac{\n 125}{6}[/tex]
⇒ [tex]k=0.8047[/tex]
Now plugging this k value in any of the equation we can find the initial population.
a)
Then,
[tex]P_0(e^0^.^8^0^4^7^\times ^2) = 400[/tex]
[tex]P_0=80.003[/tex]
The initial population of the bacteria culture is 80.003
b)
Expression for the population after [tex]t[/tex] hours.
[tex]P(t)=80.003\times e^0^.^8^0^4^7 ^(^t^)[/tex]
So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).
