We find the value of N₀ since we are provided with initial conditions.
The condition is that, at time t = 0, the amount of substance contains originally 10 grams.
We substitute:
10 = N₀ (e^(-0.1356)*0)
10 = N₀ (e^0)
N₀ = 10
When the substance is in half-life (meaning, the half of the original amount), it contains 5 grams. We solve t in this case.
5 = 10 e^(-0.1356*t)
0.5 = e^(-0.1356*t)
Multiply natural logarithms on both sides to bring down t.
ln(0.5) = -0.1356*t
Hence,
t = -(ln(0.5))/0.1356
t ≈ 5.11 days (ANSWER)
