Question

Bacteria Culturea. A certain culture of the bacterium Streptococcus A initially has 10 bacteria and is observed to bacteria in the culture after t hours. b. (b) Estimate the number of bacteria after 35 hours.c) Whenwillthebacteriacountreach10,000

Answer 1

Answer:

a. $n = 10(2^{(t/1.5)})$ b. 1.054 × 10⁸ bacteria. c. 14.95 hours

Step-by-step explanation:

Here is the complete question

Bacteria Culture A certain culture of the bacterium Streptococcus  initially has 10 bacteria and is observed to double every 1.5 hours.

(a) Find an exponential model  for the number of bacteria in the culture after  hours.

(b) Estimate the number of bacteria after 35 hours.

(c) When will the bacteria count reach  10,000.

Solution

Let n be the number of bacterium at time t. The differential equation representing the rate of bacterial growth n' with the initial number of bacteria is n' = kn.

Solving this we have,

n'/n = k

∫n'/n = ∫k

㏑n = kt + c

$n = e^{kt + c} = e^{kt} e^{c} = Ae^{kt}$

at initial conditions, t = 0 and n = 10

$10 = Ae^{k X 0} = Ae^{0} = A\\A = 10\\n = 10e^{kt}$

Since the bacteria doubles every 1.5 hours, at n = 2 × 10 = 20 at t = 1.5.

So,

$20 = 10e^{k1.5}\\\frac{20}{10} = e^{k1.5}\\2 = e^{1.5k}$

㏑2 = 1.5k

k = ln2/1.5

$n = 10e^{1.5k}$

n = 10e^(t㏑2/1.5) = 10[e^㏑2)](t/1.5)

$n = 10(2^{(t/1.5)})$

b. When t = 35 hours,

$n = 10(2^{(t/1.5)}) = 10(2^{(35/1.5)}) = 10(2^{(23.333)})$

n = 10 × 1.0544 × 10⁷ = 1.0544 × 10⁸ = 105445924.49

n ≅ 1.054 × 10⁸ bacteria.

c. We find t when n = 10,000

$10000 = 10(2^{(t/1.5)})\\\frac{10000}{10} = (2^{(t/1.5)})\\1000 = (2^{(t/1.5)})$

taking natural logarithm of both sides

㏑1000 = ㏑2^(t/1.5)

㏑1000 = t㏑2/1.5

t = 1.5㏑1000/㏑2 = 1.5 × 9.9656 = 14.95 hours

t ≅ 15 hours