Chapter 3.5, Problem 35E

a.

To determine

To calculate : the population of Antioch increasing or decreasing from $2006$ through $2012$

Answer to Problem 35E

The population of Antioch increasing

Explanation of Solution

Given information : The question is

The population of Antioch, California, from $2006$ through $2012$ can be modeled by $P=90e^{0.013t}$ , where $t$ is year , with $t=6$

The population of Antioch increasing because the model is in the form of $y=a{e}^{bx}$ where the exponent is positive which is an exponential growth model

b.

To determine

To calculate : the population in $2006,2009\&2012$

Answer to Problem 35E

The population in $2006$ is $97301$ , in $2009$ is $101171$ , in year $2012$ is $105194$

Explanation of Solution

Given information : The question is

The population of Antioch, California, from $2006$ through $2012$ can be modeled by $P=90e^{0.013t}$ , where $t$ is year , with $t=6$

Calculation :

In year $2006$

  $\begin{array}{l}t=6\left(2006\right)\\ P=90e^{0.013t}\\ P=90e^{0.013\left(6\right)}\\ P\approx 97.301\end{array}$

In year $2009$

  $\begin{array}{l}t=9\left(2009\right)\\ P=90e^{0.013t}\\ P=90e^{0.013\left(9\right)}\\ P\approx 101.171\end{array}$

In year $2012$

  $\begin{array}{l}t=12\left(2012\right)\\ P=90e^{0.013t}\\ P=90e^{0.013\left(12\right)}\\ P\approx 105.194\end{array}$

c.

To determine

When will the population of Antioch be approximately $116,000$

Answer to Problem 35E

In $2019$ the population of Antioch be approximately $116,000$

Explanation of Solution

Given information : The question is

The population of Antioch, California, from $2006$ through $2012$ can be modeled by $P=90e^{0.013t}$ , where $t$ is year , with $t=6$

Calculation :

  $\begin{array}{l}P=116\left(116000\right)\\ 116=90e^{0.013t}\\ \frac{58}{45}=e^{0.013t}\\ \ln \frac{58}{45}=\ln e^{0.013t}\\ \ln \frac{58}{45}=0.013t\\ t=\frac{\ln \frac{58}{45}}{0.013t}\\ t\approx 19.5\end{array}$