Question

5. A type of bacteria doubles in number every 25 minutes. Find the constant k

for this type of bacteria, then write the equation for modeling this exponential
growth.

Answer 1

To find the constant k for this type of bacteria, we can use the formula for exponential growth:

N(t) = N0 * e^(kt)

where N(t) is the number of bacteria at time t, N0 is the initial number of bacteria, e is Euler's number (approximately 2.71828), and k is the constant we're looking for.

We know that the bacteria doubles in number every 25 minutes, which means that after 25 minutes, the number of bacteria will be 2 times the initial number (N0). Therefore, we can write:

N(25) = 2 * N0

Substituting this into the formula, we get:

2 * N0 = N0 * e^(k*25)

Dividing both sides by N0 and simplifying, we get:

2 = e^(25k)

Taking the natural logarithm of both sides, we get:

ln(2) = 25k

Solving for k, we get:

k = ln(2)/25 ≈ 0.0278

Therefore, the equation for modeling the exponential growth of this type of bacteria is:

N(t) = N0 * e^(0.0278t)

*IG:whis.sama_ent

Answer 2

The constant k is the growth rate per unit of time. In this case, the bacteria double every 25 minutes, so we can calculate the growth rate as follows:

k = ln(2)/25

where ln(2) is the natural logarithm of 2.

To model the exponential growth of the bacteria population over time, we can use the equation:

N(t) = N0 * e^(kt)

where N(t) is the population size at time t, N0 is the initial population size, e is the mathematical constant approximately equal to 2.71828, k is the growth rate constant we just calculated, and t is the time elapsed.

So, if we start with an initial population of N0 bacteria, the population after time t can be calculated as:

N(t) = N0 * e^(kt)