Chapter 3.5, Problem 34E

To determine

To find: number of bacteria in 6 hours.

Answer to Problem 34E

397

Explanation of Solution

Given:

The initial population is 250 bacteria, and the population after 10 hours is double the population after 1 hour.

Concept Used:

Exponential growth model: $y=a{e}^{bx}$ , $b>0$

Let y denotes the number of bacteria after t hours.

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

  $y=a{e}^{bx}$ ….. (1)

Also, from given information, when $x=0$ , $y=250$ Substitute this information in the above model.

  $\begin{array}{l}250=a{e}^{b\left(0\right)}\\ 250=a{e}^0\\ 250=a\left(1\right)\\ 250=a\end{array}$

Substitute this value in equation (1),

  $y=250e^{bx}$ ….. (2)

Population after 1 hour is found by substituting $x=1$ in equation (2).

  $y=250e^b$

Population after 10 hour is found by substituting $x=10$ in equation (2).

  $y=250e^{10b}$

Now, it is given that, the population after 10 hours is double the population after 1 hour

So, this gives that,

  $\begin{array}{c}250e^{10b}=2\times 250e^b\\ e^{10b}=2e^b\\ e^{9b}=2\end{array}$

Take natural log on both sides,

  $\begin{array}{c}\ln e^{9b}=\ln 2\\ 9b\ln e=\ln 2\\ 9b=\ln 2\\ b=\frac{1}{9}\ln 2\end{array}$

Substitute in equation (2),

  $y=250e^{\left(\frac{1}{9}\ln 2\right)x}$ ….. (3)

Now, to find number of bacteria in 6 hours substitute $x=6$ in equation (3),

  $\begin{array}{c}y=250e^{\left(\frac{1}{9}\ln 2\right)\times 6}\\ \approx 397\end{array}$

Conclusion:

After 6 hours number of bacteria will be approximately 397.