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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

Best Answer 

 #1
avatar+26718 
+10

Use N = 56ek*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 = 56ek*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 = 56e0.007455*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
avatar+26718 
+10
Best Answer

Use N = 56ek*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 = 56ek*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 = 56e0.007455*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

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