A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 2 hours, there are 4,800 bacteria. At the end of 4 hours, there are 19,00 bacteria. How many bacteria were present initially?
A = P*e^kt
4800 = P*e^2k
19000 = p*e^4k
Now, 4800/p = e^2k, so
19000 = p*(4800/p)^2
19000 = 4800^2/p
p = 4800^2/19000 = 1212
just for grins, what's k?
4800=1212*e^2k
e^2k = 3.958
2k = ln 3.958 = 1.376
k = 0.688
so, A(x) = 1212*e^.688t
Bacteria experiment. If after one hour there were 1600 bacteria. Three hours later there was 400 bacteria. How many bacteria were there originally?
To determine the initial number of bacteria, we can use the exponential growth formula:
N = N0 * (1 + r)^t
Where:
N = final number of bacteria
N0 = initial number of bacteria
r = growth rate
t = time (in this case, in hours)
We are given the following information:
At the end of 2 hours, there are 4,800 bacteria.
N = 4,800
t = 2
At the end of 4 hours, there are 19,000 bacteria.
N = 19,000
t = 4
We can set up two equations using the given information:
4,800 = N0 * (1 + r)^2
19,000 = N0 * (1 + r)^4
Now we can solve these equations simultaneously to find the values of N0 and r.
To find out the initial number of bacteria, we need to use the exponential growth formula, which relates the final population size (N) to the initial size (N0), the growth rate (r), and the time (t):
N = N0 * e^(rt)
In this case, we have two data points: N = 4800 after 2 hours and N = 19000 after 4 hours. By substituting these values into the equation, we can set up a system of equations and solve for N0 and r.
First, let's plug in the values for the first data point:
4800 = N0 * e^(2r)
Next, let's plug in the values for the second data point:
19000 = N0 * e^(4r)
Now, we have a system of two equations with two unknowns (N0 and r). To find the value of N0, we can divide the second equation by the first equation:
19000 / 4800 = (N0 * e^(4r)) / (N0 * e^(2r))
3.96 = e^(2r)
To solve for r, we can take the natural logarithm of both sides:
ln(3.96) = ln(e^(2r))
ln(3.96) = 2r
Now, we can divide both sides by 2:
ln(3.96) / 2 = r
Using a calculator, we find that r ≈ 0.470.
Now that we have the value of r, we can substitute it back into either equation to solve for N0. Let's use the first equation:
4800 = N0 * e^(2(0.470))
Simplifying:
4800 = N0 * e^0.94
Dividing both sides by e^0.94:
4800 / e^0.94 = N0
Using a calculator, we find that N0 ≈ 1111.
Therefore, the initial number of bacteria is approximately 1,111.