At the beginning of a study, a certain culture of bacteria has a population of 340. The population grows according to a continuous exponential growth model. After 11 days, there are 544 bacteria.
(a) Lett be the time (in days) since the beginning of the study, and let y be the number of bacteria at time i. x 5 ? Write a formula relating y tot. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
y = 10:
(b) How many bacteria are there 23 days after the beginning of the study? Do not round any intermediate computations, and round your answer to the nearest whole number.
bacteria

Answer:

y = 340e^0.0427276t ; 908

Step-by-step explanation:

Given that :

t = time in days since study began

Number of bacteria after 11 days = 544

y = number of bacteria at time t

Exponential growth model:

y = ae^rt

Where a = initial amount ; r = Growth rate

y = 544 ; t = 11 ; a = 340

544 = 340*e^11t

544/340 = e^11t

1.6 = e^11t

Take the In of both sides

In(1.6) = 11t

0.4700036 = 11t

t = 0.4700036 / 11

t = 0.0427276

y = 340e^0.0427276t

Number of bacteria after 23 days :

Using the formula above :

t = 23

y = 340e^0.0427276(23)

y = 340 * 2.6717529

y = 908.39600

y = 908

The required formula is [tex]y = 340e^{0.0427}[/tex]

908 bacteria are there 23 days after the beginning of the study .

Given that,

A certain culture of bacteria has a population of 340.

The population grows according to a continuous exponential growth model. After 11 days, there are 544 bacteria.

We have to find,

Write a formula relating y tot. Use exact expressions to fill in the missing parts of the formula.

How many bacteria are there 23 days after the beginning of the study.

According to the question,

And let y be the number of bacteria at time,

t = time in days since study began

Number of bacteria after 11 days = 544

y = number of bacteria at time t

Then,

Exponential growth model:

[tex]y = a.e^{rt}[/tex]

Where, a = initial amount ; r = Growth rate

y = 544 ; t = 11 ; a = 340

Then,

[tex]544 = 340.e^{11t}\\\\\dfrac{544}{340} = e^{11t}\\\\1.6 = e^{11t}\\\\Taking \ log \ on \ both \ sides,\\\\ln(1.6) = 11t\\\\0.47 = 11t \\\\t = \dfrac{0.47}{11}\\\\t = 0.0427[/tex]

The required formula is [tex]y = 340e^{0.0427}[/tex]

[tex]y = 340e^{0.0427}[/tex]

Where, t = 23

[tex]y = 340e^{0.0427276(23)}\\\\y = 340 \times 2.6717529\\\\y = 908.39600\\\\y = 908[/tex]

Therefore, 908 bacteria are there 23 days after the beginning of the study .

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