Chapter 3.4, Problem 85E

To determine

To find: the year in which population exceeds 720 thousand

Answer to Problem 85E

2011

Explanation of Solution

Given:

The population P (in thousands) in the years 2005 through 2015 can be modeled by

  $P=75\ln t+540$ , $5\le t\le 15$ ….. (1)

where t represents the year, with $t=5$ corresponding to 2005.

To find the year in which population exceeds 720 thousand, substitute $P=720$ in equation (1),

  $\begin{array}{c}720=75\ln t+540\\ 720-540=75\ln t\\ 180=75\ln t\\ \frac{180}{75}=\ln t\end{array}$

Exponentiate both sides,

  $e^{\ln t}=e^{\frac{180}{75}}$

Use $e^{\ln x}=x$ ,

  $\begin{array}{c}t=e^{\frac{180}{75}}\\ t\approx 11.02\end{array}$

Hence, the population exceeds 720 thousand during the year 2011.