Acountry's population in 1993 was 127 million. in 2000 it was 132 million. estimate the population in 2008 using the exponential growth formula. round your answer to the nearest million.

To solve the exponential growth application question we proceed as follows;
suppose the time, t between 1993 to 2000 is such that in 1993, t=0 and in 2000, t=7.
Note theta the population is in millions;
The exponential formula is given by:
f(t)=ae^(kt)
where;
f(t) =current value
a=initial value
k=constant of proportionality
t=time
substituting the values we have in our formula we get:
132=127e^(7k)
132/127=e^(7k)
introducing the natural logs we get:
ln (132/127)=7k
k=[ln(132/127)]/7
k=0.0055
Thus our formula will be:
f(t)=127e^(0.0055t)
The population in 2008 will be:
f(t)=127e^(15*0.0055)=127e^(0.0825)=137.922
Thus the population in 2008 is appropriately 138 million.

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