Suppose that the number of bacteria in a certain population increases according to a continuous exponential growth model. A sample of 1000
bacteria selected from this population reached the size of 1274
bacteria in five hours. Find the hourly growth rate parameter.
Note: This is a continuous exponential growth model.
Write your answer as a percentage. Do not round any intermediate computations, and round your percentage to the nearest hundredth.
1000 e^(kt) = 1274
e^(5k) = 1.274
take ln of both sides
5k = ln 1.274
k = ln 1.274/5
= .048323...
= 4.83 % rounded to the nearest hundreth
To find the hourly growth rate parameter, we can use the formula for continuous exponential growth:
N(t) = N0 * e^(kt)
Where:
N(t) is the size of the population at time t
N0 is the initial size of the population
e is the base of the natural logarithm (approximately 2.71828)
k is the growth rate parameter
t is the time in hours
We are given that the initial size of the population (N0) is 1000 bacteria, and after 5 hours, the population size (N(t)) is 1274 bacteria.
Let's substitute these values into the formula:
1274 = 1000 * e^(5k)
To solve for k, we need to isolate it on one side of the equation. Divide both sides of the equation by 1000:
1.274 = e^(5k)
Now, take the natural logarithm of both sides of the equation. This will remove the exponential:
ln(1.274) = ln(e^(5k))
Using the properties of logarithms, we can bring down the exponent:
ln(1.274) = 5k * ln(e)
Since ln(e) equals 1, we can simplify further:
ln(1.274) = 5k
Now, divide both sides of the equation by 5 to isolate k:
k = ln(1.274) / 5
Using a calculator, we can find the value of ln(1.274) and then divide by 5:
k ≈ 0.0488
So, the hourly growth rate parameter, k, is approximately 0.0488. To express this as a percentage, multiply by 100:
0.0488 * 100 ≈ 4.88
Therefore, the hourly growth rate parameter is approximately 4.88%.