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)
