Question

The formula A0.0466129emodels the population of a particular city, in thousands, t years after 1996.When will the population of the city reach 164 thousand?1990200120031991

Answer 1

Since the equation of population is

$A=129e^{0.046t}$

Where A is the population of thousands after the year 1996

We need to find the year that has a population of 164 thousand

Then substitute A by 164

$164=129e^{0.046t}$

Divide both sides by 129

$\begin{gathered} \frac{164}{129}=\frac{129e^{0.046t}}{129} \\ \frac{16}{129}=e^{0.046t} \end{gathered}$

Insert ln on both sides

$ln(\frac{164}{129})=ln(e^{0.046t})$

Use the rule of ln to simplify

$ln(e^m)=m$$ln(\frac{164}{129})=0.046t$

Divide both sides by 0.046 to find the value of t

$\begin{gathered} \frac{ln(\frac{164}{129})}{0.046}=\frac{0.046t}{0.046} \\ 5.218565727=t \\ 5\approx t \end{gathered}$

Then the population would be 164 thousand after about 5 years after 1996

Then to find the year add 5 to 1996

$1996+5=2001$

The population will be 164 thousand on 2001

The answer is 2001