The population of a particular city doubled from 1900 to 1960. If the rate of natural increase of the population at any time is proportional to the population at the time and the population in 1960 was 60,000, estimate the population in the year 2020?
Attempt solution:
$y(0)$= the population in $1900$ = half the population in $ 1960 = 30,000$
$y(60)$ = the population in $1960 = 60,000$
$$y(t) = y_0 e^{kt}$$ $$ y(60) = 30,000 e^{60k} $$ $$ 60,000 = 30,000 e^{60k} $$ $$ 20,000 = e^{60k} $$ $$ ln(20,000) = 60k $$ $$ k = \frac{ln(20,000)}{60}$$
The population in 2020, or after 120 years is $$ y(120) = 30,000 e^{120k} $$
Which is equal to $1.19 $ x $ 10^{13}$
It seems that the result is not plausible so, what should be the correct solution and answer?