A certain type of bacteria is growing at an exponential rate that can be modeled by the equation y=ae^kt, where t represents the number of hours. There are 300 bacteria initially, and 1500 bacteria 5 hours later.
How many bacteria will be present at time t=10? Find you answer to the nearest whole bacteria. Show all steps using properties of logarithms
Determine the time at which the number of bacteria reaches 12,500. Round your answer to the nearest hour.
find k using
300 e^(5k) = 1500
use that k with t=10
finally, use that k to find t when
300 e^(kt) = 12500
t= 7508.4 i think
I dont know how to do the second part
clearly t= 7508.4 i think is not a reasonable answer. 7500 years?
300 e^(5k) = 1500
e^(5k) = 5
5k = ln5
k = ln5/5 = 0.32
so,
y(t) = 300 e^(0.32t)
y(10) = 300 e^3.2 = 7360
300 e^(0.32t) = 12500
e^(0.32t) = 41.67
0.32t = ln 41.67 = 3.73
t = 3.73/0.32 = 11.65
Y=ae^-kt
1500=300e^k4
ln5=e^k4 ln
1.609=-k4
*divide by 4 and by -1 to get rid of the negative
O.4024=k
To find the number of bacteria at time t = 10, we can plug in the given values into the equation.
Given:
Initial number of bacteria (y₀) = 300
Number of bacteria after 5 hours (y₁) = 1500
The equation for the growth of bacteria is:
y = ae^kt
Let's substitute the given values into the equation:
300 = ae^(5k) ----- (1)
1500 = ae^(10k) ----- (2)
Now, solve equations (1) and (2) simultaneously to find the values of a and k:
Divide equation (2) by equation (1):
1500 / 300 = (ae^(10k)) / (ae^(5k))
Simplify:
5 = e^(10k - 5k)
5 = e^(5k)
Now, take the natural logarithm (ln) of both sides:
ln(5) = ln(e^(5k))
ln(5) = 5k
Solve for k:
k = ln(5) / 5
Substitute the value of k into equation (1) to find a:
300 = ae^(5 * (ln(5)/5))
300 = ae^(ln(5))
300 = a * 5
a = 300/5 = 60
So, the equation for the given data is:
y = 60e^(kt)
Now, substitute t = 10 into the equation:
y = 60e^(10k)
Calculate k from earlier:
k = ln(5) / 5 ≈ 0.322
Then, substitute k and t = 10 into the equation to find the number of bacteria at t = 10:
y = 60e^(0.322 * 10)
y = 60e^(3.22)
y ≈ 60 * 25.09072 ≈ 1505.34
Therefore, the estimated number of bacteria at time t = 10 is about 1505.
To find the time at which the number of bacteria reaches 12,500, we can substitute the given values into the equation:
12,500 = 60e^(kt)
Divide both sides by 60:
12,500 / 60 = e^(kt)
Calculate ln(12,500 / 60):
ln(12,500 / 60) = kt
Solve for t:
t = ln(12,500 / 60) / k
Using the earlier value of k ≈ 0.322:
t ≈ ln(12,500 / 60) / 0.322
Evaluate the expression to find t:
t ≈ ln(208.3333) / 0.322 ≈ 4.04
Rounded to the nearest hour, the time it takes for the number of bacteria to reach 12,500 is approximately 4 hours.