This exercise uses the population growth model. It is observed that a certain bacteria culture has a relative growth rate of 19% per hour, but in the presence of an antibiotic the relative growth rate is reduced to 7% per hour. The initial number of bacteria in the culture is 23. Find the projected population after 24 hours for the following conditions. (Round your answers to the nearest whole number.) (a) No antibiotic is present, so the relative growth rate is 19%. 105 Incorrect: Your answer is incorrect. bacteria (b) An antibiotic is present in the culture, so the relative growth rate is reduced to 7%. 39 Incorrect: Your answer is incorrect. bacteria

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.

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