A bacteria culture grows with constant relative growth rate. The bacteria count was 2,560 after 4 hours and 2,621,440 after 12 hours.(a) What is the

A) [tex]P = P_0 e^{rt}[/tex]

B) [tex]19564[/tex]

C) [tex]5.5[/tex] or [tex]6[/tex] hours

Explanation:

A) The formula that can be used for bacteria after "t" hours is

[tex]P = P_0 e^{rt}[/tex]

Where ,

[tex]P[/tex] represents the population after "t" hours

[tex]P_0[/tex] represents the initial population.

[tex]r[/tex] represents the rate of growth of population

[tex]t[/tex] represents the time taken by the population to grow

B) Here let us first find the rate of population growth.

[tex]9000 = 400 * e^{r*4}\\3.11 = r* 4\\r = 0.778[/tex]

Thus, the population of bacteria after 5 hours

[tex]P = 400 * e^{0.778 * 5} \\P = 400* 48.91\\P = 19564\\[/tex]

C)

[tex]30000 = 400 * e^{0.778 * t) \\\\4.317 = 0.778 * t\\t = 5.549\\or t= 5.5 or 6[/tex]

A) The expression for the number of bacteria is [tex]P(t) = 400e^{0.7783t}[/tex].

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

[tex]P' = kP[/tex]

where [tex]P[/tex] stands for the population of bacteria.

Writing [tex]P'[/tex] as [tex]\frac{dP}{dt}[/tex], we get

[tex]\frac{dP}{dt} = kP[/tex].

Notice that this is a separable equation, so

[tex]\frac{dP}{P} = kdt[/tex].

Then, integrating in both sides of the equality:

[tex]\int\frac{dP}{P} = \int kdt[/tex].

We have,

[tex]\ln P = kt+C[/tex].

Now, taking exponential

[tex]P(t) = Ce^{kt}[/tex].

The next step is to find the value for the constant [tex]C[/tex]. We do this using the initial condition [tex]P(0)=400[/tex]. Recall that this is the initial population of bacteria. So,

[tex]400 = P(0) = Ce^{k0}=C[/tex].

Hence, the expression becomes

[tex]P(t) = 400e^{kt}[/tex].

Now, we find the value for [tex]k[/tex]. We are going to use that [tex]P(4)=9000[/tex]. Notice that

[tex]9000 = 400e^{k4}[/tex].

Then,

[tex]\frac{90}{4} = e^{4k}[/tex].

Taking logarithm

[tex]\ln\frac{90}{4} = 4k[/tex], so [tex]\frac{1}{4}\ln\frac{90}{4} = k[/tex].

So, [tex]k=0.7783788273[/tex], and approximating to the fourth decimal place we can take [tex]k=0.7783[/tex]. Hence,

[tex]P(t) = 400e^{0.7783t}[/tex].

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

[tex]P(5) =400e^{0.7783*5} = 19593.723 \approx 19593[/tex].  

C) In this case we want to do the reverse operation: we want to find the value of t such that

[tex]30000 = 400e^{0.7783t}[/tex].

This expression is equivalent to

[tex]75 = e^{0.7783t}[/tex].

Now, taking logarithm we have

[tex]\ln 75 = 0.7783t[/tex].

Finally,

[tex]t = \frac{\ln 75}{0.7783} \approx 5.55[/tex].

So, after 5.55 hours the population of bacteria will reach 30000.

A. 86.64%

B. 80

C. y(t)= 80[tex]e^{0.8664t}[/tex]

D. y(5.5) = 9388

E. y(t) = 8134 bacteria for every hour

F.  8.28 hours

Step-by-step explanation:

a. The bacteria count after 4 hours y(4) = 2560

The bacteria count after 12 hours y(12) = 2621440

equation 1: ye^4k = 2560

equation 2: ye^12k= 2621440

divide (2) by (1),

e^8k = 1024

Taking ln both sides:

ln(e8k ) = ln(1024)

8k = 6.931

k=0.8664      // after dividing both sides by 8

rate of growth k% = 0.8664*100 = 86.64%

b. substituting for k in equation 2:

   ye^4*0.8664 = 2560

   Initial bacteria count y(4) = 80

c. y(t)= 80e^0.8664t

d. the number of bacteria after 5.5 hours => y(5.5) = 80[tex]e^{0.8664(5.5)}[/tex] ≈ 9388

e. y'(5.5) = 80*0.8664e^0.8664(5.5) = 69.31e^0.8664(5.5) = 8134

f. for bacteria population = 104000

where ye^0.8664t = 104000

   y = 80

   80e^0.8664t = 104000    //divide both sides by 80

   e^0.8664t = 1300

   ln(e^0.8664t = ln(1300)

   t = 7.17 / 0.8664 = 8.28 hours