1. **State the problem:** We need to find an exponential regression equation of the form $$y = ab^x$$ that best fits the given data points: (2,162), (3,163), (4,146), (5,139), (9,111), (12,92), (16,71). 2. **Formula and explanation:** The exponential regression model is $$y = ab^x$$ where $a$ is the initial value (when $x=0$) and $b$ is the base or growth/decay factor. 3. **Transform the data:** Take the natural logarithm of both sides to linearize: $$\ln y = \ln a + x \ln b$$ Let $Y = \ln y$, $A = \ln a$, and $B = \ln b$, so the equation becomes: $$Y = A + Bx$$ This is a linear equation in terms of $x$ and $Y$. 4. **Calculate $Y = \ln y$ for each $y$:** - $\ln 162 \approx 5.0876$ - $\ln 163 \approx 5.0938$ - $\ln 146 \approx 4.9836$ - $\ln 139 \approx 4.9345$ - $\ln 111 \approx 4.7095$ - $\ln 92 \approx 4.5218$ - $\ln 71 \approx 4.2627$ 5. **Find the means:** $$\bar{x} = \frac{2+3+4+5+9+12+16}{7} = \frac{51}{7} \approx 7.286$$ $$\bar{Y} = \frac{5.0876 + 5.0938 + 4.9836 + 4.9345 + 4.7095 + 4.5218 + 4.2627}{7} \approx 4.7857$$ 6. **Calculate slope $B$:** $$B = \frac{\sum (x_i - \bar{x})(Y_i - \bar{Y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$(2-7.286)(5.0876-4.7857) + (3-7.286)(5.0938-4.7857) + \ldots + (16-7.286)(4.2627-4.7857) \approx -8.927$$ Calculate denominator: $$(2-7.286)^2 + (3-7.286)^2 + \ldots + (16-7.286)^2 \approx 110.857$$ So, $$B = \frac{-8.927}{110.857} \approx -0.0805$$ 7. **Calculate intercept $A$:** $$A = \bar{Y} - B \bar{x} = 4.7857 - (-0.0805)(7.286) = 4.7857 + 0.5865 = 5.3722$$ 8. **Convert back to $a$ and $b$:** $$a = e^A = e^{5.3722} \approx 215.5$$ $$b = e^B = e^{-0.0805} \approx 0.923$$ 9. **Write the exponential regression equation:** $$y = 215.5 \times 0.923^x$$ 10. **Round coefficients to the nearest thousandth:** $$y = 215.500 \times 0.923^x$$ This equation models the data with exponential decay as $x$ increases.