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

Need help with this problem. It's taken from the chapter "Logarithm and Polynomials-example-1
User Zachvac
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1 Answer

5 votes

Answer:

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 )

------------------------------------------------------------------------

(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

-----------------------------------------------------------------------

(c)

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

User Gokhan Tank
by
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