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]
