1. **State the problem:** We are given data points $(50,64.7)$, $(60,51.3)$, $(70,40.5)$, $(90,25.9)$, and $(100,7.5)$ and a model $XY^A = B$. We want to find the best linear fit using the least squares method. 2. **Rewrite the model:** The model is $XY^A = B$. Taking natural logarithms on both sides gives: $$\ln(X) + A \ln(Y) = \ln(B)$$ Rearranged as: $$A \ln(Y) = \ln(B) - \ln(X)$$ Or equivalently: $$\ln(Y) = \frac{1}{A} (\ln(B) - \ln(X))$$ 3. **Linearize the model:** Let $u = \ln(X)$ and $v = \ln(Y)$. The equation becomes: $$v = c + m u$$ where $m = -\frac{1}{A}$ and $c = \frac{\ln(B)}{A}$. 4. **Calculate $u$ and $v$ for each data point:** - For $(50,64.7)$: $u=\ln(50) \approx 3.912$, $v=\ln(64.7) \approx 4.170$ - For $(60,51.3)$: $u=\ln(60) \approx 4.094$, $v=\ln(51.3) \approx 3.937$ - For $(70,40.5)$: $u=\ln(70) \approx 4.248$, $v=\ln(40.5) \approx 3.700$ - For $(90,25.9)$: $u=\ln(90) \approx 4.499$, $v=\ln(25.9) \approx 3.255$ - For $(100,7.5)$: $u=\ln(100) \approx 4.605$, $v=\ln(7.5) \approx 2.015$ 5. **Apply least squares formulas:** Calculate sums: $$\sum u = 3.912 + 4.094 + 4.248 + 4.499 + 4.605 = 21.358$$ $$\sum v = 4.170 + 3.937 + 3.700 + 3.255 + 2.015 = 17.077$$ $$\sum u v = (3.912)(4.170) + (4.094)(3.937) + (4.248)(3.700) + (4.499)(3.255) + (4.605)(2.015) = 16.31 + 16.11 + 15.72 + 14.65 + 9.28 = 72.06$$ $$\sum u^2 = 3.912^2 + 4.094^2 + 4.248^2 + 4.499^2 + 4.605^2 = 15.30 + 16.76 + 18.05 + 20.24 + 21.21 = 91.56$$ Number of points $n=5$. 6. **Calculate slope $m$ and intercept $c$:** $$m = \frac{n \sum u v - \sum u \sum v}{n \sum u^2 - (\sum u)^2} = \frac{5 \times 72.06 - 21.358 \times 17.077}{5 \times 91.56 - (21.358)^2} = \frac{360.3 - 364.8}{457.8 - 456.0} = \frac{-4.5}{1.8} = -2.5$$ $$c = \frac{\sum v - m \sum u}{n} = \frac{17.077 - (-2.5)(21.358)}{5} = \frac{17.077 + 53.395}{5} = \frac{70.472}{5} = 14.094$$ 7. **Find $A$ and $B$:** Recall $m = -\frac{1}{A}$ so: $$A = -\frac{1}{m} = -\frac{1}{-2.5} = 0.4$$ And $c = \frac{\ln(B)}{A}$ so: $$\ln(B) = A c = 0.4 \times 14.094 = 5.638$$ $$B = e^{5.638} \approx 282.3$$ 8. **Final model:** $$XY^{0.4} = 282.3$$ Or equivalently: $$Y = \left(\frac{282.3}{X}\right)^{\frac{1}{0.4}} = \left(\frac{282.3}{X}\right)^{2.5}$$ This is the least squares linear fit for the given data and model.