Question

A quantity P obeys the exponential growth law P(t)=e5t (t in years).

At what time t0 is P=15? Use yr for years. t0=?
At what time t1 is P=20? Use yr for years. t1= ?
What is the doubling time for P? Use yr for years. The doubling time is ?

Answer 1

the time at which P =15 is
`{t_o = 0.54161 \ years}`

the time at which P =20 is
`{t_1 = 0.59914 \ years}`

the doubling time
`{t = 0.138629 \ years}`

Explanation:

From the information we are being provided with:

The quantity P i.e
`P(t) = e ^(5t)`

In order to determine the time at which P = 15, we have the following:

`t= t_o` when P = 15

∴
`e^(5t_o) = 15`

`5t_o= In (15)`

`t_o = (1)/(5)In 15`

`t_o = (1)/(5) * 2.70805`

`{t_o = 0.54161 \ years}`

Hence, the time at which P =15 is
`{t_o = 0.54161 \ years}`

In order to determine the time at which P = 20, we have the following:

`t= t_1` when P = 20

∴
`e^(5t_1) = 20`

`5t_1= In (20)`

`t_1= (1)/(5)In (20)`

`t_1 = (1)/(5) * 2.9957`

`{t_1 = 0.59914 \ years}`

Hence, the time at which P =20 is
`{t_1 = 0.59914 \ years}`

The doubling time at t = 0, mean P = 2

∴
`e^(5t) = 2`

`5t= In (2)`

`t= (1)/(5)In(2)`

`t = (1)/(5) *0.693147`

`{t = 0.138629 \ years}`

Hence, the doubling time
`{t = 0.138629 \ years}`