a countrys population in 1995 was 56 million.in 2002 it was 59 million.estimate the population in 2016 using exponential growth formula round to the nearest million.?

Guest Dec 10, 2014

#1**+10 **

Use N = 56e^{k*t} where t is the time in years from 1995 and k is an. as yet, unknown rate constant.

Find k from the 2002 data: 59 = 56e^{k*7} (because 2002 - 1995 = 7)

Divide through by 56, take logs, then divide by 7 to get k = ln(59/56)/7

$${\mathtt{k}} = {\frac{{ln}{\left({\frac{{\mathtt{59}}}{{\mathtt{56}}}}\right)}}{{\mathtt{7}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.007\: \!455\: \!107\: \!595\: \!795\: \!7}}$$

In 2016 t = 2016 - 1995 = 21, so

N = 56e^{0.007455*21}

$${\mathtt{N}} = {\mathtt{56}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.007\: \!455}}{\mathtt{\,\times\,}}{\mathtt{21}}\right)} \Rightarrow {\mathtt{N}} = {\mathtt{65.490\: \!604\: \!574\: \!065\: \!260\: \!8}}$$

or N ≈ 65 million

.

Alan
Dec 10, 2014

#1**+10 **

Best Answer

Use N = 56e^{k*t} where t is the time in years from 1995 and k is an. as yet, unknown rate constant.

Find k from the 2002 data: 59 = 56e^{k*7} (because 2002 - 1995 = 7)

Divide through by 56, take logs, then divide by 7 to get k = ln(59/56)/7

$${\mathtt{k}} = {\frac{{ln}{\left({\frac{{\mathtt{59}}}{{\mathtt{56}}}}\right)}}{{\mathtt{7}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.007\: \!455\: \!107\: \!595\: \!795\: \!7}}$$

In 2016 t = 2016 - 1995 = 21, so

N = 56e^{0.007455*21}

$${\mathtt{N}} = {\mathtt{56}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.007\: \!455}}{\mathtt{\,\times\,}}{\mathtt{21}}\right)} \Rightarrow {\mathtt{N}} = {\mathtt{65.490\: \!604\: \!574\: \!065\: \!260\: \!8}}$$

or N ≈ 65 million

.

Alan
Dec 10, 2014