1. **Problem statement:** We have data points $(x,y)$ and want to fit an exponential model of the form $$y = Ck^{dx}$$ to the data. 2. **Goal:** Find parameters $C$, $k$, and $d$ to fit the model, then predict $y$ at $x_0 = 12.31$ using two methods: - (1) Direct substitution into the model: $$y = Ck^{d x_0}$$ - (2) Using the last observed point $(x_{10}=10, y_{10}=508)$ and relative change to estimate $y$ at $x_0$. 3. **Step 1: Linearize the model.** Taking natural logarithm on both sides: $$\ln y = \ln C + d x \ln k$$ Let $A = \ln C$ and $B = d \ln k$, then: $$\ln y = A + B x$$ 4. **Step 2: Compute $\ln y$ for each data point:** $$(1,3) \to \ln 3 = 1.0986$$ $$(2,4) \to \ln 4 = 1.3863$$ $$(3,1) \to \ln 1 = 0$$ $$(4,7) \to \ln 7 = 1.9459$$ $$(5,14) \to \ln 14 = 2.6391$$ $$(6,35) \to \ln 35 = 3.5553$$ $$(7,62) \to \ln 62 = 4.1271$$ $$(8,126) \to \ln 126 = 4.8363$$ $$(9,259) \to \ln 259 = 5.5568$$ $$(10,508) \to \ln 508 = 6.2304$$ 5. **Step 3: Perform linear regression on $(x, \ln y)$ to find $A$ and $B$:** Calculate means: $$\bar{x} = \frac{1+2+\cdots+10}{10} = 5.5$$ $$\overline{\ln y} = \frac{1.0986 + 1.3863 + 0 + 1.9459 + 2.6391 + 3.5553 + 4.1271 + 4.8363 + 5.5568 + 6.2304}{10} = 3.3376$$ Calculate slope $B$: $$B = \frac{\sum (x_i - \bar{x})(\ln y_i - \overline{\ln y})}{\sum (x_i - \bar{x})^2} = \frac{54.818}{82.5} = 0.6644$$ Calculate intercept $A$: $$A = \overline{\ln y} - B \bar{x} = 3.3376 - 0.6644 \times 5.5 = -0.3206$$ 6. **Step 4: Recover parameters:** $$C = e^A = e^{-0.3206} = 0.7259$$ Since $B = d \ln k$, we can set $d=1$ for simplicity, then: $$\ln k = B = 0.6644 \implies k = e^{0.6644} = 1.9437$$ 7. **Step 5: Predict $y$ at $x_0=12.31$ using method (1):** $$y = C k^{d x_0} = 0.7259 \times 1.9437^{12.31}$$ Calculate exponent: $$1.9437^{12.31} = e^{12.31 \times 0.6644} = e^{8.179} = 3583.7$$ So: $$y = 0.7259 \times 3583.7 = 2601.5$$ 8. **Step 6: Predict $y$ at $x_0=12.31$ using method (2):** Use last point $(10, 508)$ and relative change per unit $x$: $$\text{Relative change per unit} = k^d = 1.9437$$ Time difference: $$\Delta x = 12.31 - 10 = 2.31$$ Predict: $$y = y_{10} \times (k^d)^{\Delta x} = 508 \times 1.9437^{2.31}$$ Calculate: $$1.9437^{2.31} = e^{2.31 \times 0.6644} = e^{1.534} = 4.64$$ So: $$y = 508 \times 4.64 = 2357.1$$ 9. **Step 7: Calculate percentage difference between methods:** $$\text{Difference} = \frac{|2601.5 - 2357.1|}{2601.5} \times 100 = \frac{244.4}{2601.5} \times 100 = 9.39\%$$ **Final answers:** - Method (1) prediction: $2601.5$ - Method (2) prediction: $2357.1$ - Percentage difference: $9.39\%$