Question

Hey there, need some help here

Answer 1

k ≈ 0.34

Explanation:

Using the law of logarithms

log
`x^(n)` ⇔ nlogx

given N =
`N_(0)`
`e^(kt)`

Where
`N_(0)` is the initial amount

here N = 7500, t = 10 and
`N_(0)` = 250, hence

7500 = 250
`e^(10k)` ( divide both sides by 250 )

`e^(10k)` = 30 ( take the ln of both sides )

ln
`e^(10k)` = ln 30

10k lne = ln 30 [ lne = 1 ]

k =
`(ln30)/(10)` ≈ 0.34