2. A bacteria population grows exponentially with growth rate k=2.5 how long it would take for the bacteria population to double its size?
2 = e^2.5t
Ln(2) = lne^2.5t
ln(2) =2.5t
t = ln(2)/.25
To determine how long it would take for the bacteria population to double its size, we need to use the formula for exponential growth:
N = N0 * e^(kt)
Where:
N is the final population size
N0 is the initial population size
k is the growth rate
t is the time in which the population grows
In this case, we know that the growth rate, k, is 2.5. To find the time it takes for the population to double, we need to find the value of t when N = 2 * N0.
Let's substitute these values into the formula:
2 * N0 = N0 * e^(2.5t)
Simplifying the equation:
2 = e^(2.5t)
To solve for t, we need to take the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^(2.5t))
Using the property of logarithms that ln(e^x) = x:
ln(2) = 2.5t
Now, we can solve for t by dividing both sides of the equation by 2.5:
t = ln(2) / 2.5
Using a calculator, we find that ln(2) is approximately 0.693. Dividing this by 2.5, we get:
t ≈ 0.693 / 2.5 ≈ 0.277
Therefore, it would take approximately 0.277 time units for the bacteria population to double its size.