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A country's population in 1995 was 228 million. In 2001 it was 230 million. Estimate the population in 2010 using the exponential growth formula. Round your answer to the nearest million.

[tex]\[ P = A e^{kt} \][/tex]

Enter the correct answer:


Sagot :

To estimate the population in 2010, we use the exponential growth formula:

[tex]\[ P = A e^{kt} \][/tex]

Step-by-step solution:

1. Identify the initial values:

- Initial population in 1995 ([tex]\( P_{1995} \)[/tex]) = 228 million
- Population in 2001 ([tex]\( P_{2001} \)[/tex]) = 230 million
- Time difference from 1995 to 2001 ([tex]\( t_{1995 \to 2001} \)[/tex]) = 2001 - 1995 = 6 years

2. Determine the growth rate constant [tex]\( k \)[/tex]:

We know that in 2001 the population is given by:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot t_{1995 \to 2001}} \][/tex]
Plugging in the known values:
[tex]\[ 230 = 228 e^{6k} \][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
[tex]\[ \ln \left(\frac{230}{228}\right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left(\frac{230}{228}\right)}{6} \][/tex]

Evaluating this, we get:
[tex]\[ k \approx 0.001456 \][/tex]

3. Estimate the population in 2010:

Time difference from 1995 to 2010 ([tex]\( t_{1995 \to 2010} \)[/tex]) = 2010 - 1995 = 15 years

Using the exponential growth formula again:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot t_{1995 \to 2010}} \][/tex]
Plugging in the values:
[tex]\[ P_{2010} = 228 e^{0.001456 \cdot 15} \][/tex]

Evaluating the exponent:
[tex]\[ P_{2010} \approx 233.033 \][/tex]

4. Round the population to the nearest million:

[tex]\[ \text{Population in 2010} \approx 233 \text{ million} \][/tex]

So, the estimated population in 2010 is 233 million.
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