Question

A 45 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1109.

Answer 1

The decay formula is
N(t) = N₀ exp(-kt)
where
N₀ = 45 g, original mass
N(t) = mass after time t, days
k = 0.1109, the decay constant.

At half life, N(t) = (1/2)*N₀. Therefore
1/2 = exp(-0.1109t)
Take natural log of each side.
ln(0.5) = -0.1109t

t = ln(0.5)/(-0.1109) = 6.25 days

Answer: 6.3 days (nearest tenth)

Answer 2

The decay formula is
N(t) = N₀ exp(-kt)
where
N₀ = 45 g, original mass
N(t) = mass after time t, days
k = 0.1109, the decay constant.

At half life, N(t) = (1/2)*N₀. Therefore
1/2 = exp(-0.1109t)
Take natural log of each side.
ln(0.5) = -0.1109t

t = ln(0.5)/(-0.1109) = 6.25 days

Answer: 6.3 days (nearest tenth)