Question

A country population in 1991 was 231 million in 1999 it was 233 million . Estimate the population in 2003 using the exponential growth formula. Round you answer to the nearest million

Answer 1

p(y)=ir^t

233=231r^(1999-1991)

(233/231)^(1/8)=r

p(y)=231(233/231)^((y-1991)/8)  so in 2003

p(2003)=231(233/231)^((2003-1991)/8)

p(2003)=231(233/231)^(1.5)

p(2003)=234

So 234 million (to the nearest million people)

Answer 2

Answer:

population in 2003 is 234 million.

Step-by-step explanation:

A country's population in 1991 was 231 million

In 1999 it was 233 million.

We have to calculate the population in 2003.

Since population growth is always represented by exponential function.

It is represented by $P(t)=P_{0}e^{kt}$

Here t is time in years, k is the growth constant, and  is initial population.

For year 1991 ⇒

233 = $P_{0}e^{8k}$ = 231 $e^{8k}$

$\frac{231}{233}= e^{8k}$

Taking ln on both the sides ⇒

$ln(\frac{233}{231})=lne^{8k}$

ln 233 - ln 231 = 8k  [since ln e = 1 ]

5.451 - 5.4424 = 8k

k = $\frac{0.0086}{8}=0.001075$

For year 2003 ⇒

$P(t)=P_{0}e^{kt}$

P (t) = 231 × $e^{(0.001075)(12)}$

     = 231 × $e^{0.0129}$

     = 231 × 1.0129

     = 233.9 ≈ 234 million

Therefore, population in 2003 is 234 million.