Question

A certain type of bacteria is growing at an exponential rate that is modeled by the equation y= ae^kt, where t represents the number of hours. There are 100 bacteria initially, and 500 bacteria five hours later.

Answer 1

Answer:

The number of bacteria after t hours is given by: $y(t) = 100e^{0.3219t}$

Step-by-step explanation:

Amount of bacteria after t hours:

Is given by the following equation:

$y(t) = ae^{kt}$

In which a is the initial value and k is the constant of growth.

There are 100 bacteria initially

This means that $a = 100$. So

$y(t) = ae^{kt}$

$y(t) = 100e^{kt}$

500 bacteria five hours later.

This means that $y(5) = 500$. We use this to find k. So

$y(t) = 100e^{kt}$

$500 = 100e^{5k}$

$e^{5k} = 5$

$\ln{e^{5k}} = \ln{5}$

$5k = \ln{5}$

$k = \frac{\ln{5}}{5}$

$k = 0.3219$

So

$y(t) = 100e^{kt}$

$y(t) = 100e^{0.3219t}$

The number of bacteria after t hours is given by: $y(t) = 100e^{0.3219t}$