Answer: Â The population of the country will be 106 millions in 2014.
Step-by-step explanation:
The exercise gives you the following exponential model, which describes the​ population "A" (in​ millions) of a country "t" years after 2003:
[tex]A=104.8 e^{0.001 t}[/tex]
In this case you must determine when the population of that country will be 106 millions, so you can identify that:
[tex]A=106[/tex]
Now you need to substitute this value into the exponential model given in the exercise:
[tex]106=104.8 e^{0.001 t}[/tex]
Finally, you must solve for "t", but first it is important to remember the following Properties of logarithms:
[tex]ln(a)^b=b*ln(a)\\\\ln(e)=1[/tex]
Then:
[tex]\frac{106}{104.8}=e^{0.001 t}\\\\ln(\frac{106}{104.8})=ln(e)^{0.001 t}\\\\ln(\frac{106}{104.8})=0.001 t(1)\\\\\frac{ln(\frac{106}{104.8})}{0.001}}=t\\\\t=11.38\\\\t\approx11[/tex]
Notice that in 11 years the population will be 106 millions, then the year will be:
[tex]2003+11=2014[/tex]
The population of the country will be 106 millions in 2014.
