1. **Problem Statement:** We have bacterial population data over time and assume it follows an exponential growth model $$y = a e^{b x}$$ where $y$ is the population in millions and $x$ is time in hours. 2. **Write the normal equations:** To linearize, take natural logarithm on both sides: $$\ln y = \ln a + b x$$ Let $Y = \ln y$ and $A = \ln a$, so the model becomes: $$Y = A + b x$$ The normal equations for least squares fitting are: $$\sum Y = n A + b \sum x$$ $$\sum x Y = A \sum x + b \sum x^2$$ where $n$ is the number of data points. 3. **Calculate sums from data:** Given data: $x = [0, 0.5, 1, 1.5, 2, 2.5]$ $y = [0.10, 0.45, 2.15, 9.15, 40.35, 180.75]$ Calculate $Y = \ln y$: $Y = [\ln 0.10, \ln 0.45, \ln 2.15, \ln 9.15, \ln 40.35, \ln 180.75] \approx [-2.3026, -0.7985, 0.7655, 2.213, 3.698, 5.198]$ Calculate sums: $$\sum x = 0 + 0.5 + 1 + 1.5 + 2 + 2.5 = 7.5$$ $$\sum x^2 = 0^2 + 0.5^2 + 1^2 + 1.5^2 + 2^2 + 2.5^2 = 0 + 0.25 + 1 + 2.25 + 4 + 6.25 = 13.75$$ $$\sum Y = -2.3026 - 0.7985 + 0.7655 + 2.213 + 3.698 + 5.198 = 8.7734$$ $$\sum x Y = 0*(-2.3026) + 0.5*(-0.7985) + 1*0.7655 + 1.5*2.213 + 2*3.698 + 2.5*5.198 = 0 - 0.3993 + 0.7655 + 3.3195 + 7.396 + 12.995 = 24.0767$$ 4. **Solve normal equations:** $$8.7734 = 6 A + 7.5 b$$ $$24.0767 = 7.5 A + 13.75 b$$ Multiply first equation by 7.5: $$8.7734 \times 7.5 = 6 \times 7.5 A + 7.5 \times 7.5 b$$ $$65.8005 = 45 A + 56.25 b$$ Multiply second equation by 6: $$24.0767 \times 6 = 7.5 \times 6 A + 13.75 \times 6 b$$ $$144.4602 = 45 A + 82.5 b$$ Subtract first from second: $$144.4602 - 65.8005 = (45 A - 45 A) + (82.5 b - 56.25 b)$$ $$78.6597 = 26.25 b$$ $$b = \frac{78.6597}{26.25} \approx 2.996$$ Substitute $b$ back into first equation: $$8.7734 = 6 A + 7.5 \times 2.996$$ $$8.7734 = 6 A + 22.47$$ $$6 A = 8.7734 - 22.47 = -13.6966$$ $$A = \frac{-13.6966}{6} = -2.2828$$ 5. **Find $a$:** $$a = e^A = e^{-2.2828} \approx 0.102$$ 6. **Final fitted model:** $$y = 0.102 e^{2.996 x}$$ 7. **Estimate population at $x=3$ hours:** $$y(3) = 0.102 e^{2.996 \times 3} = 0.102 e^{8.988}$$ Calculate $e^{8.988} \approx 8,000$ (approximate) $$y(3) \approx 0.102 \times 8000 = 816$$ So the estimated population at 3 hours is approximately 816 million. **Answer:** The bacterial population at 3 hours is approximately **816 million**.