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

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

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 = (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)`