The number of bacteria in a culture is increasing according to the law of exponential growth. There are 125 bacteria in the culture after 2 hours and 350 bacteria after 4 hours.
a) Find the initial population.
b) Write an exponential growth model for the bacteria population. Let t represent time in hours.
c) Use the model to determine the number of bacteria after 8 hours.
d) After how many hours will the bacteria count be 25,000?
Thank you!!
let the number be n
n = a e^(kt), where a is the initial number and t is the time in hours, k is a constant
case 1, when t=2, n = 125
125 = a e^2k
case 2 , when t = 4, n= 350
350 = a e^4k
divide the two equations
350/125 = a e^4k / (a e^(2k))
2.8 = e^2k
2k = ln 2.8
k = .5148
in first equation,
125 = a e^(2(.5148))
a = 44.64
a) so initially there were 46 bacteria.
b) n = 44.64 e^(.5148t)
c) when t = 8 , n = 44.64 e^(8(.5148)) = appr. 2744
d) 25000 = 44.64 e^(.5148 t)
560.0358 = e^ .5148t
.5148t = ln 560.0358
t = 12.29 hours
a) Well, if we want to find the initial population of bacteria, we can assume the bacteria count at time 0 hours. So, let's call that "P₀" for initial population.
b) To write an exponential growth model, we can use the formula P(t) = P₀ * e^(kt), where P(t) represents the population at time t, P₀ is the initial population, e is Euler's number (approximately 2.71828), and k is the growth rate constant.
c) Using the given information, we can set up two equations:
At t = 2 hours, P(2) = 125, which gives us the equation 125 = P₀ * e^(2k).
And at t = 4 hours, P(4) = 350, giving us 350 = P₀ * e^(4k).
Now, we have a system of equations to solve for P₀ and k. You can solve these equations by substituting one into the other or by taking the logarithm of both sides. But don't worry, I'll do it for you.
After solving the system of equations, we find that P₀ is approximately 50 and k is approximately 0.3466.
d) Now that we have our exponential growth model, we can use it to find the number of bacteria after 8 hours. Plugging in the values, we get P(8) ≈ 50 * e^(0.3466 * 8).
And after some math, we find that the number of bacteria after 8 hours is approximately 1917.
e) Lastly, we want to know after how many hours the bacteria count will be 25,000. To find this, we can rearrange our exponential growth model and solve for t:
25,000 = 50 * e^(0.3466t).
Now, divide both sides by 50 and take the natural logarithm of both sides to isolate t:
ln(25,000/50) = 0.3466t.
By using a calculator or the power of math, we find that t ≈ 14.219 hours.
Remember, these calculations are based on the assumption that bacteria just love to multiply exponentially and don't have any external limitations. Happy counting!
a) To find the initial population, we can use the formula for exponential growth:
N(t) = N0 * e^(kt)
where N(t) is the population at time t, N0 is the initial population, e is Euler's number (approximately 2.71828), k is the growth rate constant, and t is the time in hours.
We are given two sets of data points: (2, 125) and (4, 350). Plugging these values into the equation, we get:
125 = N0 * e^(2k) ---(1)
350 = N0 * e^(4k) ---(2)
Next, let's solve these equations simultaneously to find the value of N0.
Dividing equation (2) by equation (1):
(350 / 125) = (N0 * e^(4k)) / (N0 * e^(2k))
2.8 = e^(2k)
Taking the natural logarithm of both sides:
ln(2.8) = ln(e^(2k))
ln(2.8) = 2k
k = ln(2.8) / 2
k ≈ 0.4812
Substituting k back into equation (1):
125 = N0 * e^(2*0.4812)
125 = N0 * e^(0.9624)
Evaluating e^(0.9624):
125 = N0 * 2.6185
Dividing both sides by 2.6185:
N0 ≈ 47.7
Thus, the initial population (N0) is approximately 47.7.
b) The exponential growth model for the bacteria population is:
N(t) = N0 * e^(kt)
Substituting the values we found before:
N(t) ≈ 47.7 * e^(0.4812t)
c) To determine the number of bacteria after 8 hours, we can use the exponential growth model obtained in part b:
N(8) ≈ 47.7 * e^(0.4812*8)
N(8) ≈ 47.7 * e^(3.8496)
Evaluating e^(3.8496):
N(8) ≈ 47.7 * 46.97
N(8) ≈ 2236.2
Thus, the number of bacteria after 8 hours is approximately 2236.2.
d) To find out after how many hours the bacteria count will be 25,000, we can use the exponential growth model again:
25,000 = 47.7 * e^(0.4812t)
Dividing both sides by 47.7:
25,000 / 47.7 = e^(0.4812t)
525.35 = e^(0.4812t)
Taking the natural logarithm of both sides:
ln(525.35) = ln(e^(0.4812t))
ln(525.35) = 0.4812t
t = ln(525.35) / 0.4812
t ≈ 8.694
Thus, it will take approximately 8.694 hours for the bacteria count to reach 25,000.
a) To find the initial population, we need to use the given information and solve for the initial population. Let's denote the initial population as P0.
We are given that after 2 hours, there are 125 bacteria. So, we can write the equation:
P = P0 * e^(k * 2)
125 = P0 * e^(k * 2) -- equation 1
Similarly, after 4 hours, the population is 350:
P = P0 * e^(k * 4)
350 = P0 * e^(k * 4) -- equation 2
We have a system of equations with two variables (P0 and k), so we need to solve these equations simultaneously to find the values of P0 and k.
To solve these equations, we can rewrite them as:
125 = P0 * e^(2k)
350 = P0 * e^(4k)
Now, we can divide equation 2 by equation 1 to eliminate P0:
350/125 = (P0 * e^(4k)) / (P0 * e^(2k))
2.8 = e^(4k - 2k)
2.8 = e^2k
Take the natural logarithm (ln) of both sides:
ln(2.8) = ln(e^2k)
ln(2.8) = 2k * ln(e)
ln(2.8) = 2k
Now, we can solve for k:
k = ln(2.8) / 2
Using a calculator, we find k ≈ 0.5878.
Now, substitute the value of k back into equation 1:
125 = P0 * e^(0.5878 * 2)
125 = P0 * e^(1.1756)
Divide both sides by e^(1.1756):
125 / e^(1.1756) ≈ P0
Using a calculator, we find P0 ≈ 81.91.
Therefore, the initial population (P0) is approximately 81.91 bacteria.
b) The exponential growth model for the bacteria population can be written as:
P(t) = P0 * e^(kt)
Where P(t) is the population at time t, P0 is the initial population, k is the growth constant.
c) Using the exponential growth model, we can determine the number of bacteria after 8 hours:
P(8) = P0 * e^(k * 8)
Substituting the values we found earlier, we have:
P(8) = 81.91 * e^(0.5878 * 8)
Using a calculator, we find P(8) ≈ 1905.56.
Therefore, the number of bacteria after 8 hours is approximately 1905.56.
d) To find after how many hours the bacteria count will be 25,000, we need to use the exponential growth model and solve for time (t).
25,000 = P0 * e^(k * t)
Substituting the known values of P0 and k, we have:
25,000 = 81.91 * e^(0.5878 * t)
Divide both sides by 81.91:
25,000 / 81.91 = e^(0.5878 * t)
Take the natural logarithm (ln) of both sides:
ln(25,000 / 81.91) = 0.5878 * t * ln(e)
ln(25,000 / 81.91) = 0.5878 * t
Now, we can solve for t:
t = ln(25,000 / 81.91) / 0.5878
Using a calculator, we find t ≈ 10.81.
Therefore, the bacteria count will be 25,000 after approximately 10.81 hours.