Explanation:
P(t) = [tex]P_{0} e^{kt}[/tex]
Population after 3 hours is 400
i.e., P(3) = 400
400 = [tex]P_{0} e^{3k}[/tex] .................(1)
Population after 10 hours is 2000 i.e. P(5) = 2000
2000 = [tex]P_{0} e^{10k}[/tex] .................(2)
From eq. (1) we get
[tex]P_{0} =400e^{-3k}[/tex]
From eq. (2) we get
[tex]P_{0} =2000e^{-10k}[/tex]
Compare above two equations and solve for k
[tex]400e^{-3k}=2000e^{-10k}[/tex]
[tex]e^{-3k}=5e^{-10k}[/tex]
By taking log on both sides, we get
- 3k = [tex]\ln (5e^{-10k})[/tex]
⇒ - 3k = [tex]\ln 5+\ln e^{-10k}[/tex]
⇒ - 3k = [tex]\ln 5-10k[/tex]
⇒ 7k = [tex]\ln 5[/tex]
⇒ k = [tex]\dfrac{\ln 5}{7}[/tex] = 0.2299
Put k = 0.2299 in equation (1), we get
400 = [tex]P_{0} e^{3k}[/tex]
⇒ 400 = [tex]P_{0} e^{3(0.2299)}[/tex]
⇒ [tex]P_{0} =\dfrac{400}{e^{3(0.2299)}}[/tex]
[tex]P_{0}[/tex] ≈ 201 bacteria
