Answer:
a) Population of bacteria in the culture without antibiotic present = 2198
b) Population of bacteria in the culture with anti biotic present = 123
Step-by-step explanation:
The growth rate of the bacteria is given as 19% per hour normally and 7% per hpur when antibiotic is present.
In differential terms,
(dN/dt) = KN
where k = growth rate and N = number of bacteria at any time t in hours.
Solving the differential equation
(dN/N) = kdt
∫ (dN/N) = ∫ kdt
Integration the left hand side from N₀ (Initial number of bacteria) to N (number of bacteria at any time) and the right hand side from 0 to t
In(N/N₀) = kt
(N/N₀) = eᵏᵗ
N = N₀ eᵏᵗ
a) So, when there is no antibiotic, rate of growth = k = 19% = 0.19
N = N₀ e⁰•¹⁹ᵗ
N₀ = 23
N = ?
t = 24 hours
0.19t = 0.19 × 24 = 4.56
N = 23 e⁴•⁵⁶ = 2198.42 = 2198 to the nearest whole number.
b) With antibiotic present, rate of growth = k = 7% = 0.07
N = N₀ e⁰•⁰⁷ᵗ
N₀ = 23
N = ?
t = 24 hours
0.07t = 0.07 × 24 = 1.68
N = 23 e¹•⁶⁸ = 123.41 = 123 to the nearest whole number.
Hope this Helps!!!
