+0

# a countrys population in 1995 was 56 million.in 2002 it was 59 million.estimate the population in 2016 using exponential growth formula roun

0
321
1

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
+26550
+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
Sort:

#1
+26550
+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

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details