1500 = 300 e^(k * 5)
ln(5) = 5 k
k = 32.2% / hr
y(t) = 300 e^(.322 t)
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.
1 answer