Question

A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.

1) Determine the exponential growth equation for this population.

2) How long will it take for the population to grow from its initial population of 250 to a population of 2000?​

Answer 1

step BY step:

To determine the exponential growth equation for the population, we can use the general form of the exponential growth model:

P(t) = P₀ * e^(rt)

Where:

P(t) is the population at time t

P₀ is the initial population

e is the base of the natural logarithm (approximately 2.71828)

r is the growth rate

t is the time

We are given that the initial population (P₀) is 250 and the population after 5 days is 1600. We can plug these values into the equation to find the growth rate (r).

1600 = 250 * e^(5r)

Divide both sides of the equation by 250:

1600/250 = e^(5r)

6.4 = e^(5r)

Taking the natural logarithm of both sides:

ln(6.4) = ln(e^(5r))

Using the property of logarithms (ln(e^x) = x):

ln(6.4) = 5r

Now, solve for r:

r = ln(6.4)/5

Using a calculator, we find that r ≈ 0.379.

Therefore, the exponential growth equation for this population is:

P(t) = 250 * e^(0.379t)

To determine how long it will take for the population to grow from 250 to 2000, we can set up the equation:

2000 = 250 * e^(0.379t)

Divide both sides of the equation by 250:

8 = e^(0.379t)

Take the natural logarithm of both sides:

ln(8) = ln(e^(0.379t))

Using the property of logarithms:

ln(8) = 0.379t

Now, solve for t:

t = ln(8)/0.379

Using a calculator, we find that t ≈ 6.034.

Therefore, it will take approximately 6.034 units of time (e.g., days, hours, etc.) for the population to grow from 250 to 2000.

Answer 2

(1) -
`P(t)=250e^(0.37126t)`

(2) -
`t \approx 5.6 \ days`

Explanation:

Answering question (1) -

If we define "P" as the population of the substance at any given time "t," then the rate at which it changes is directly proportional to the quantity of bacteria existing at that specific time.

We have,

`(dP)/(dt) \propto P`

We can rewrite this as,

`(dP)/(dt) = kP`

Where "k" is the constant of proportionality.

We can solve the differential equation using separation of variables.

`\boxed{\left\begin{array}{ccc}\text{\underline{Separable Differential Equation:}}\\(dy)/(dx) =f(x)g(y)\\\\\rightarrow\int(dy)/(g(y))=\int f(x)dx \end{array}\right }`

`(dP)/(dt) = kP\\\\\\ \\\Longrightarrow (1)/(P) dP= kdt\\\\\\\\\Longrightarrow \displaystyle\int (1)/(P) dP= \displaystyle \int kdt\\\\\\\\\Longrightarrow \boxed{\ln(P)=kt+C}`

Rearranging and solving for "P." We get,

`\Longrightarrow e^(\ln(P))=e^(kt+C)\\\\\\\\\therefore \boxed{P=Ce^(kt)}`

We need to find the value of "C," the constant of integration. We can use the following initial condition:

`\text{When } t=0, \ \text{then} \ P(0)=P_0. \ \text{Where} \ P_0 \ \text{represents the initial population. }`

`P=Ce^(kt); \ P(0)=P_0\\\\\\\\\Longrightarrow P_0=Ce^(k(0))\\\\\\\\\therefore \boxed{C=P_0}`

Thus, our general formula we will use for question (1) is,

`\rightarrow\bold{P(t)=P_0e^(kt)}`

We can now use the initial conditions given to us in the question.

Given:

`P_0=250\\\\P(5)=1600`

Using the given information to find the value of "k."

`\Longrightarrow 1600=250e^(k(5))\\\\\\\\\Longrightarrow (1600)/(250)=e^(k(5))\\\\\\\\\Longrightarrow \ln\Big((32)/(5)\Big)=5k\\\\\\\\\Longrightarrow k= (\ln\Big((32)/(5)\Big))/(5) \\\\\\\\\therefore \boxed{k \approx 0.37126}`

Therefore, the equation representing the population is as follows:

`\boxed{\boxed{P(t)=250e^(0.37126t)}}`

Answering question (2) -

To find the time it takes for the population to grow from 250 to 2000, we can use the exponential growth equation we just found and solve for "t" using the following condition:

`P(??)=2000`

Plugging in the condition.

`P(t)=250e^(0.37126t)}; \ P(??)=2000\\\\\\\\\Longrightarrow 2000=250e^(0.37126t)\\\\\\\\\Longrightarrow (2000)/(250)=e^(0.37126t)\\\\\\\\\Longrightarrow \ln(8)=0.37126t\\\\\\\\\Longrightarrow t=(\ln(8))/(0.37126)\\\\\\\\\therefore \boxed{\boxed{t \approx 5.6 \ days}}`

Thus, the problem is solved.