Question

Need help with this problem. It's taken from the chapter "Logarithm and Polynomials".

Answer 1

see explanation

Explanation:

Given

A = 80
`e^(-kt)` ← initial amount is 80

The amount is halved in 1.5 days, hence

(a)

80
`e^(-1.5k)` = 40 ( divide both sides by 80 )

`e^(-1.5k)` = 0.5

Take the natural log of both sides

ln
`e^(-1.5k)` = ln 0.5, thus

-1.5k ln e = ln 0.5 ← ln e = 1

- 1.5k = ln 0.5 ( divide both sides by - 1.5 )

k =
`(ln0.5)/(-1.5)` = 0.462 ( 3 dec. places )

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(b)

80
`e^(-0.462t)` = 5 ( divide both sides by 80 )

`e^(-0.462t)` = 0.0625

Take the natural log of both sides

ln
`e^(-0.462t)` = ln 0.0625

- 0.462t = ln 0.0625 ( divide both sides by - 0.462

t =
`(ln0.0625)/(-0.462)` ≈ 6 days

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(c)

80
`e^(-0.642(9))` = 80 ×
`e^(-4.158)` ≈ 1.25 g