The Population size of a country was 12.7 million in the year 2000 and 14.3 million in the year 2010. Assume that the population size follows an exponential function.

What will the population size be in the year 2015, assuming the formula holds until then?

In t years after 2000 the
population will be P(t) = P_{0}e^{kt}. Given P(0) =
12.7×10^{6}
and P(10) = 14.3×10^{6},
determine P_{0}
= 12.7×10^{6
}and 14.3×10^{6
}= 12.7×10^{6}e^{10k}.
From the latter equation e^{10k}
= 14.3/12.7 and

k =
ln(14.3/12.7)/10 = 0.01187. The population grows according to the
exponential law as

P(t)
= 12.7×10^{6
}e^{0.01187t}.

In
2015 the population will be P(15) = 12.7×10^{6
}e^{0.01187}^{×15}= 15.2 million.

If
P(t) = 18×10^{6},
then from the equation 18×10^{6
}= 12.7×10^{6
}e^{0.01187t}we get e^{0.01187t}
= 18/12.7 = 1.417. Take the natural logarithm of both sides: 0.01187
t = ln(1.417) and t = ln(1.417)/0.01187 = 29.4 years

The independent variable is the time (in years) that passed since 2000 and the dependent variable is the population size. The population size is changing according to the exponential law, that is, P(t) = P_0 times e^(kt), where P_0 and k are some constants that can be found if two corresponding values of t and P(t) are given. If t = 0, then e^0 = 1, so P(0) = P_0 is the initial population (in 2000).

The second pair t = 10 and P(10) = 14,300,000 was used to find the constant k. After both P_0 and k are found, it is possible to use the formula P(t) = P_0 times e^(kt) either to find P(t) for given t or to find t for given value of P(t).

or 3(2x + 1)/x >= 0. The expression on the left-hand side changes it sign at the points x = - 1/2 (where 2x + 1 = 0) and x = 0. They divide the number line into three parts

x < -1/2 there both 2x + 1 and x are negative and their quotient is positive (note that at x = -1/2 the inequality holds, 0 >= 0)

-1/2 < x < 0 there 2x + 1 > 0 and x < 0, so the quotient (2x + 1)/x < 0

x > 0 both 2x + 1 and x are negative and their quotient is positive