a bacteria pop. grows exponentially. There are 1500 bacteria after 3 hours and 20,000 after 8 hours.
ahving trouble finding the initial bacteria pop.
thanks for the help
In five hours the population get's larger by a factor 13 + 1/3----->
N(t) = N_0 * (13 + 1/3)^(t/5 hours)
Insert in here t = 3 hours and solve for N _0:
N_0 = N(3 hours)* (13 + 1/3)^(-3/5) = 317
thanks
To find the initial bacteria 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
k is the growth rate
e is the base of the natural logarithm
Given the information that there are 1500 bacteria after 3 hours and 20,000 bacteria after 8 hours, we can set up two equations with N(t) and solve for N₀ and k.
Equation 1: 1500 = N₀ * e^(3k)
Equation 2: 20000 = N₀ * e^(8k)
To solve for N₀, we can divide Equation 2 by Equation 1:
20000 / 1500 = e^(8k) / e^(3k)
Simplifying, we have:
13.33 = e^(5k)
Taking the natural logarithm of both sides to eliminate the exponential term:
ln(13.33) = ln(e^(5k))
ln(13.33) = 5k * ln(e)
ln(13.33) = 5k
Dividing by 5:
k = ln(13.33) / 5
Now that we have the value of k, we can substitute it back into either Equation 1 or Equation 2 to solve for N₀. Let's use Equation 1:
1500 = N₀ * e^(3 * ln(13.33) / 5)
Simplifying:
1500 = N₀ * e^(3 * ln(13.33))^(1/5)
1500 = N₀ * e^(ln(13.33^3))^(1/5)
1500 = N₀ * e^(ln(2187.37))^(1/5)
1500 = N₀ * e^(0.839)^(1/5)
1500 = N₀ * e^(0.1678)
Dividing by e^(0.1678):
N₀ = 1500 / e^(0.1678)
Using a calculator, we find:
N₀ ≈ 317
Therefore, the initial bacteria population is approximately 317.