Step-by-step explanation:
Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.
From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by [tex]\frac{100+1.8}{100}[/tex] = [tex]\frac{101.8}{100}[/tex].
So, the population in the year t can be given by [tex]P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}[/tex]
Population in the year 2000 = [tex]3,381,000\textrm{x}(\frac{101.8}{100})^{6}[/tex]=[tex]3,762,979.38[/tex]
Population in year 2000 = 3,762,979
Let us assume population doubles by year [tex]y[/tex].
[tex]2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}[/tex]
[tex]log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})[/tex]
[tex]y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537[/tex]
[tex]y[/tex]≈[tex]2033[/tex]
∴ By 2033, the population doubles.
