Question

 A country's population in 1991 was 48 million. In 2002 it was 51 million. estimate the population in 2008 using the exponential growth formula. Round to nearest million.

Answer 1

Using exponential growth, we get P1=P0*e^rt, where P1 is the result population, P0 is the initial population, r is the rate, and t is time. So:
51=48*e^11r
1.0625=e^11r
ln 1.0625=ln e^11r=11r ln e=11r
r=ln 1.0625/11=0.0055
Then
51*e^0.00551132925603953114369146654913*(2008-2002)=51*1.0336207977813425465939939376231=52.71466 million as the population
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Answer 2

Answer:

53 Million

Step-by-step explanation:

In 1991: t = 0 and A = 45 million

In 2002: t = 11 and p = 51 Million

You're Solving For k:

P = $Ae^{kt}$ In(1.0625) - In( $e^{k*11}$ )

51 = $48^{k*11}$ 0.0606 = k * 11

1.0625 = $e^{k*11}$ 0.0055 ≈ k

In 2008: t = 17

P = $Ae^{kt}$

P = $48e^{0.0055*17}$

P = $48e^{0.094}$

P = 48*1.0986

P ≈ 52.71466 million as the population

P ≈ 53 Million If Rounded