Chapter 3, Problem 14PS

To determine

To find: the exponential decay model that approximates the number of bacteria B in the culture after t hours.

Answer to Problem 14PS

  $B=500{\left(\frac{2}{5}\right)}^t$

Explanation of Solution

Given:

The initial population is 500 bacteria, and the population after 2 hours is 200.

Concept Used:

Exponential decay model: $B=B_0{a}^{kt}$

Let y denotes the number of bacteria after t hours.

From given information, number of bacteria follows law of exponential decay model, so is given by:

  $B=B_0{a}^{kt}$ ….. (1)

Also, from given information, when $t=0$ , $B=500$ Substitute this information in the above model.

  $\begin{array}{l}500=B_0{a}^{k\left(0\right)}\\ 500=B_0{a}^0\\ 500=B_0\left(1\right)\\ 500=B_0\end{array}$

Substitute this value in equation (1),

  $B=500a^{kt}$ ….. (2)

Also, from given information, when $t=2$ , $B=200$ Substitute this information in the equation (2),

  $\begin{array}{c}200=500a^{k\left(2\right)}\\ \frac{200}{500}=a^{2k}\\ \frac{2}{5}=a^{2k}\end{array}$

Taking square root both sides,

  $\begin{array}{c}{\left(a^{2k}\right)}^{\frac{1}{2}}={\left(\frac{2}{5}\right)}^{\frac{1}{2}}\\ a^k={\left(\frac{2}{5}\right)}^{\frac{1}{2}}\end{array}$

Substituting this value in equation (2), the required exponential decay model is:

  $B=500{\left(\frac{2}{5}\right)}^t$