Respuesta :
Solution:
Consider the exponential model
[tex]A=883.1e^{0.019t}[/tex]Where A is the population, and t is years after 2003. Then if A= 1315 million, the above equation becomes:
[tex]\text{ 1315 x 10}^6\text{ = }=883.1e^{0.019t}[/tex]solving for the exponential e, we get:
[tex]\text{ }\frac{\text{ 1315 x 10}^6}{883.1}\text{ }=e^{0.019t}[/tex]that is:
[tex]e^{0.019t}\text{ =1489072.585}[/tex]now, applying natural logarithm to both sides of the equation, we obtain:
[tex]ln(e^{0.019t})\text{ =ln(1489072.585)}[/tex]this is equivalent to:
[tex]^{}\text{ 0.019t=ln(1489072.585)}[/tex]solving for t, we get:
[tex]^{}\text{t=}\frac{\text{ln(1489072.585)}}{\text{ 0.019}}=\frac{14.21}{0.019}=748.08[/tex]that is, 748.08 years after 2003, that is, in the year:2751.08
So that, the solution is:
The population of the country will be 1316 million in the year 2751.08 or 748.08 years after 2003.
