Chapter 3.4, Problem 86E

To determine

To find: the year in which population exceeds 965 thousand

Answer to Problem 86E

2007

Explanation of Solution

Given:

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

  $P=81\ln t+807$ , $5\le t\le 15$ ….. (1)

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

Concept used:

Power property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a{u}^n=n{\log }_{a}u$

To find the year in which population exceeds 965 thousand, substitute $P=965$ in equation (1),

  $\begin{array}{c}965=81\ln t+807\\ 965-807=81\ln t\\ 158=81\ln t\\ \frac{158}{81}=\ln t\end{array}$

Exponentiate both sides,

  $e^{\ln t}=e^{\frac{158}{81}}$

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

  $\begin{array}{c}t=e^{\frac{158}{81}}\\ t\approx 7.03\end{array}$

Hence, the population exceeds 965 thousand during the year 2007.