full question is in the description box

Question

a countrys population in 1993 was 83 million in 2001 it was 87 million estimate the population in 2004 using the exponential growth formula round answer to nearest million

Answer 1

A = a₀e^kt

87 = 83e^8k

1.048 ≈ e^8k

ln 1.048 ≈ 8k

0.047 ≈ 8k

0.00586 ≈ k

A ≈ 87e^0.00586(3)

A ≈ 87e^0.018

A ≈ (87)(1.018)

 

Answer 2

P(t) = P₀·b^t

• P(t) = population at time t in millions
• P₀ = population in 1993 = 83 million
• b = growth rate
• t = years since 1993

 

P(t) = 83·b^t

 

Now we'll use the two data points they gave us, (1993,83) and (2001,87), to find the value of b:

 

For 1993, t = 1993 - 1993 = 0

For 2001, t = 2001 - 1993 = 7

 

P(0) = 83·b⁰ = 83·1 = 83

P(7) = 83·b⁷

P(7)/P(0) = b⁷

87/83 = b⁷

(87/83)^1/7 = b = 1.00675

 

P(t) = 83·(1.00675)^t

 

To find the population in 2004:

 

t = 2004-1993 = 11

P(11) = 83·(1.00675)¹¹ = 89 million (rounded to the nearest million)

 

Answer 3

Call 1993 year zero (t = 0).  Then 2001 is year 8 (t = 8 years).

 

P(t) = P₀e^kt, t in years, k in years^-1, P in millions

 

 

P₀ = 83 (at t = 0)

P(8) = 87 (at t = 8)

 

87 = 83e^k(8)

 

k = ln(87/83) / 8 ≅ 0.00588344 years^-1

 

P(t) = 83e^0.00588344t

 

2004 will be year 11 after 1993 ⇒

 

P(11) = 83e^{0.00588344(11)} ≅ 88.5

 

In 2004 the population should 88.5 million, which you may round up to 89 million.