Question

How do you find the exponential function, `n(t)=n_oe^(kt)` that satisfies the conditions `n(0)=3` and `n(7)=21`?

Answer 1

`n(t) = n_0 e^(kt)`
if `n_0 = 3` and `k = ln(7)/7`
then
`n(0) = 3 e^(ln(7)/7)(0)) = 3 e^0 = 3 (1) = 3` as required
and
`n(7) = 3 e^(ln(7)/7)(7)) = 3 e^(ln(7)) = 3 * 7 = 21` as required

How we got there :
We are given that:
`n(0) = 3` and
`n(0) = n_0 e^((k)(0))` after substituting 0 for `t`
or
`n(0) = n_0 e^0 = n_0 * (1) = n_0`
So,
`n_0 = n(0) = 3`

Now we have
`n(7) = 21` (given)
and
`n(7) = 3 * e^((k)(7))`
so
`e^((k)(7)) = 7`

Since `ln(e^p) = p`
if we take the natural log of both sides
`ln(e^((k)(7))) = ln(7)`
becomes
`(k)(7) = ln(7)`
or
`k = ln(7)/7`

and the original equation:
`n(t) = n_0 e^(kt)`
becomes
`n(t) = 3 e^(((ln(7))/7)(t))`