Question

A population of bacteria is growing according to the equation P(t)=850e0.19t Estimate when the population will exceed 1685. t=?

`e^(0.19t)`

Answer 1

`t=1.56411 sec`

Explanation:

Given:
`P=850e^(0.19t)`

`P=1685`

`t=?`

`1685=850e^(0.19t)`

`1685/850=e^(0.19t)`

`1.982353=e^(0.19t)`

Taking logarithms

`ln1.982353=0.19t`

`0.297181=0.19t`

`0.19t=0.297181`

`t=0.297181/0.19`

`t=1.56411 sec`

Answer 2

The population will exceed 1685 when t = 3.5

Explanation:

Given : P(t) = `850e^(20.19t)`
To find: When P = 1685 t = ?
Solution: P(t) = `850e^(0.19t)`
1685 = `850e^(0.19t)`
`1685/850`= `e^(0.19t)`
1.98 = `e^(0.19t)`
`ln 1.98` = 0.19t
0.68 = 0.19t
t = 3.6
( you haven't mentioned about the unit of time )