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.
Find the rate of growth, k, of the bacteria.
Write the equation that can be model the growth of the bacteria at time, t.
1500 = 300 e^(k * 5)
ln(5) = 5 k
k = 32.2% / hr
y(t) = 300 e^(.322 t)
To find the rate of growth, k, of the bacteria, we can use the given information and the exponential growth model equation.
Given:
Initially, there are 300 bacteria, which corresponds to y = 300.
After 5 hours, there are 1500 bacteria, which corresponds to y = 1500 when t = 5.
Using these values, we can plug them into the exponential growth equation y = ae^kt to form a system of equations:
For the initial condition: y = 300, t = 0
300 = ae^k*0
300 = a * e^0
300 = a
For the condition after 5 hours: y = 1500, t = 5
1500 = ae^k*5
1500 = a * e^5k
Now we have a system of equations:
a = 300
1500 = 300 * e^5k
Divide the second equation by the first equation:
(1500 / 300) = (300 / 300) * (e^5k)
5 = e^5k
To solve for k, take the natural log of both sides to eliminate the exponential:
ln(5) = ln(e^5k)
ln(5) = 5k * ln(e)
ln(5) = 5k * 1
ln(5) = 5k
So, the rate of growth, k, can be found by taking the natural logarithm of 5 and dividing it by 5.
Now let's write the equation that models the growth of the bacteria at any given time, t.
The equation is y = ae^kt, where:
y represents the number of bacteria at time t,
a represents the initial number of bacteria,
e is the base of natural logarithms, and
k represents the rate of growth of the bacteria.
In this case, since a = 300 and we have determined k, we can write the equation as:
y = 300e^(k * t)