Question

The amount of phosphorus present in a sample at a given time is given by the following equation:

Qt = Q0e-0.03t

What percent of the initial amount of phosphorus is left after 25 days of decay?

Answer 1

$Q_t = Q_0e^{-0.03t} \\ \frac{Q_t}{Q_0} = e^{-0.03 \times 25} \\ \frac{Q_t}{Q_0} = e^{-0.75} \\ \frac{Q_t}{Q_0} = e^{-0.75} \\ \frac{Q_t}{Q_0} = 0.472$

Proportion:

Q₀ – 100%

Q_t – x

$x = \frac{Q_t \times 100}{Q_0} \\ = 0.472 \times 100 = 47.2$ %

Answer: 47.2 %