Exponential growth and decay

Many problem sums have equations that include exponential terms to model growth and decay.

Example: Population growth

Question

The population of a bacteria culture is modeled by $P = 100 e^{0.5 t},$ where $t$ is time in hours.

Find the initial population.
Find the population after 4 hours.
How long does it take for the population to double?

Solution

When $t = 0$ : $\begin{matrix} P & = 100 e^{0} \\ = 100 ∎ \end{matrix}$

When $t = 4$ : $\begin{matrix} P & = 100 e^{0.5 (4)} \\ = 100 e^{2} \\ \approx 738.9 ∎ \end{matrix}$

The population doubles when $P = 200.$ $\begin{matrix} 200 & = 100 e^{0.5 t} \\ 2 & = e^{0.5 t} \\ \ln 2 & = 0.5 t \\ t & = \frac{\ln 2}{0.5} \\ \approx 1.39 hours ∎ \end{matrix}$

Example: Radioactive decay

Question

A radioactive substance decays according to the model $M = 500 e^{- 0.04 t},$ where $M$ is the mass (in grams) remaining after $t$ years.

Find the initial mass.
Find the mass remaining after 20 years, giving your answer to 3 significant figures.
How long does it take for the mass to reduce to 100 g? Give your answer to 3 significant figures.

Solution

When $t = 0$ : $\begin{matrix} M & = 500 e^{0} \\ = 500 g ∎ \end{matrix}$

When $t = 20$ : $\begin{matrix} M & = 500 e^{- 0.04 (20)} \\ = 500 e^{- 0.8} \\ \approx 225 g ∎ \end{matrix}$

Set $M = 100$ and solve for $t$ : $\begin{matrix} 100 & = 500 e^{- 0.04 t} \\ 0.2 & = e^{- 0.04 t} \\ \ln 0.2 & = - 0.04 t \\ t & = \frac{\ln 0.2}{- 0.04} \\ \approx 40.2 years ∎ \end{matrix}$