1. **Problem Statement:** Find the value of the constant $k$ in the first table where $x$ and $f(x)$ values are given as: $$\begin{array}{c|c} x & f(x) \\\hline 0.3 & 2 \\ 3 & 5 \\ 30 & 8 \\ k & 11 \\ 3000 & 14 \end{array}$$ 2. **Understanding the problem:** The function $f(x)$ appears to be logarithmic since the values increase slowly and the $x$ values increase multiplicatively. 3. **Formula and rules:** For a logarithmic function of the form $$f(x) = a \log_b(x) + c,$$ where $a$, $b$, and $c$ are constants, the function values increase by a constant amount when $x$ is multiplied by a fixed factor. 4. **Check the pattern:** Observe the increments in $f(x)$ as $x$ increases: - From $0.3$ to $3$ (multiplied by 10), $f(x)$ increases from 2 to 5 (increase of 3). - From $3$ to $30$ (multiplied by 10), $f(x)$ increases from 5 to 8 (increase of 3). - From $30$ to $3000$ (multiplied by 100), $f(x)$ increases from 8 to 14 (increase of 6). This suggests $f(x)$ is proportional to $\log_{10}(x)$, scaled and shifted. 5. **Express $f(x)$ in terms of $\log_{10}(x)$:** Since each multiplication by 10 increases $f(x)$ by 3, $$f(x) = 3 \log_{10}(x) + d,$$ where $d$ is a constant. 6. **Find $d$ using a known point:** Use $x=0.3$, $f(0.3)=2$: $$2 = 3 \log_{10}(0.3) + d$$ Calculate $\log_{10}(0.3)$: $$\log_{10}(0.3) = \log_{10}\left(\frac{3}{10}\right) = \log_{10}(3) - 1 \approx 0.4771 - 1 = -0.5229$$ So, $$2 = 3(-0.5229) + d = -1.5687 + d \implies d = 2 + 1.5687 = 3.5687$$ 7. **Final function:** $$f(x) = 3 \log_{10}(x) + 3.5687$$ 8. **Find $k$ such that $f(k) = 11$:** $$11 = 3 \log_{10}(k) + 3.5687$$ Subtract 3.5687: $$11 - 3.5687 = 3 \log_{10}(k)$$ $$7.4313 = 3 \log_{10}(k)$$ Divide both sides by 3: $$\log_{10}(k) = \frac{7.4313}{3} = 2.4771$$ Convert from log to exponential form: $$k = 10^{2.4771}$$ Calculate: $$k \approx 10^{2.4771} = 10^{2} \times 10^{0.4771} = 100 \times 3 = 300$$ 9. **Answer:** $$\boxed{k = 300}$$ This matches the pattern of the table where $x$ values increase by factors of 10 and $f(x)$ increases by 3.