Increased/Decreased Populations

Let:
- x be the initial population
- y be the final population
- z be the number of years passed since January 1st 2000

We know 1.02 is the rate of change (obtained from the 2% growth) so from here we create a standard rate of change expression.

y = x (1.02)^{ z }

Plugging in the values we know. 2,000,000 as y and 4 as z

2000000= x (1.02)^{4}

2000000=1.08243216 x

2000000/1.08243216= x

• Whilst rounding to nearest whole number:
  2000000/1.08243216= 1847691

• Therefore:
  x =1847691
  x =1848000 (to the nearest thousand)

Use the standard compound interest forumla;

$2,000,000 \times {0.98}^{3} \approx 1,848,000$

(0.98 for the 2% decrease going backwards each year and the power of three as we are going back 3 years)

let x be the population of the country 2002 then by increase of 2% we have x + 2% x = 2000000 by solving we get 1960784 similarly repeating the process to 2000 on the fixed rate of 2 % we get 1846000