Sure! Let's solve this problem step by step using the exponential growth formula [tex]\( P = A e^{kt} \)[/tex].
### Step 1: Understand the given information
- Population in 1995 [tex]\((P_{1995})\)[/tex]: 228 million
- Population in 2001 [tex]\((P_{2001})\)[/tex]: 230 million
- Year difference from 1995 to 2001 [tex]\((\Delta t_{1995-2001})\)[/tex]: [tex]\(2001 - 1995 = 6\)[/tex] years
- Year difference from 1995 to 2010 [tex]\((\Delta t_{1995-2010})\)[/tex]: [tex]\(2010 - 1995 = 15\)[/tex] years
### Step 2: Use the exponential growth formula
The exponential growth formula is:
[tex]\[ P = A e^{kt} \][/tex]
We need to find the growth rate [tex]\( k \)[/tex]. To do this, we use the populations in 1995 and 2001:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot \Delta t_{1995-2001}} \][/tex]
[tex]\[ 230 = 228 e^{6k} \][/tex]
### Step 3: Solve for [tex]\( k \)[/tex]
First, divide both sides by 228:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
Take the natural logarithm of both sides:
[tex]\[ \ln \left( \frac{230}{228} \right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left( \frac{230}{228} \right)}{6} \][/tex]
Using the natural logarithm:
[tex]\[ k \approx \frac{0.0088}{6} \approx 0.0015 \][/tex]
### Step 4: Use [tex]\( k \)[/tex] to estimate the population in 2010
We use the value of [tex]\( k \)[/tex] we found to estimate the population in 2010:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot \Delta t_{1995-2010}} \][/tex]
[tex]\[ P_{2010} = 228 e^{0.0015 \cdot 15} \][/tex]
Calculate the exponent:
[tex]\[ 0.0015 \cdot 15 = 0.0225 \][/tex]
So,
[tex]\[ P_{2010} = 228 e^{0.0225} \][/tex]
Calculate the exponential term:
[tex]\[ e^{0.0225} \approx 1.0228 \][/tex]
Finally, calculate [tex]\( P_{2010} \)[/tex]:
[tex]\[ P_{2010} \approx 228 \cdot 1.0228 \approx 233.224 \][/tex]
Rounded to the nearest million, the population in 2010 is [tex]\( 233 \)[/tex] million.
### Conclusion
The estimated population in 2010 is:
[tex]\[
\boxed{233}
\][/tex]
