Question:

Step 1: Rewrite the inequality using logarithms. We start with the inequality $\left(\dfrac{1}{x}\right)^{\lambda/x} \leq \dfrac{1}{9}$ and take the natural logarithm of both sides: $$\dfrac{\lambda}{x} \ln\left(\dfrac{1}{x}\right) \leq \ln\left(\dfrac{1}{9}\right)$$ Step 2: Simplify using logarithm properties. Using the property $\ln\left(\dfrac{1}{a}\right) = -\ln a$: $$\dfrac{\lambda}{x} \cdot (-\ln x) \leq -\ln 9$$ $$-\dfrac{\lambda \ln x}{x} \leq -\ln 9$$ Step 3: Isolate the condition on $\lambda$. Multiplying both sides by $-1$ (and reversing the inequality): $$\dfrac{\lambda \ln x}{x} \geq \ln 9$$ Therefore, for a positive integer $x$ to be a solution: $$\lambda \geq \dfrac{x \ln 9}{\ln x}$$ Step 4: Test positive integer values of $x$. We need to find which positive integer $x$ gives the minimum value of $\dfrac{x \ln 9}{\ln x}$. **For $x = 1$:** $\ln 1 = 0$, which causes division by zero. This case is undefined, so we skip it. **For $x = 2$:** $$\dfrac{2 \ln 9}{\ln 2} = \dfrac{2 \times 2.197}{0.693} = \dfrac{4.394}{0.693} \approx 6.34$$ So $\lambda \geq 6.34$, meaning $\lambda \geq 7$ for integer $\lambda$. **For $x = 3$:** $$\dfrac{3 \ln 9}{\ln 3} = \dfrac{3 \times 2.197}{1.099} = \dfrac{6.591}{1.099} \approx 6.0$$ So $\lambda \geq 6$ for integer $\lambda$. **For $x = 4$:** $$\dfrac{4 \ln 9}{\ln 4} = \dfrac{4 \times 2.197}{1.386} = \dfrac{8.788}{1.386} \approx 6.34$$ So $\lambda \geq 6.34$, meaning $\lambda \geq 7$ for integer $\lambda$. **For $x = 9$:** $$\dfrac{9 \ln 9}{\ln 9} = 9$$ So $\lambda \geq 9$ for integer $\lambda$. Step 5: Identify the minimum value of $\lambda$. From the calculations above, the minimum requirement is $\lambda \geq 6$, which occurs at $x = 3$. Step 6: Verify the solution. For $x = 3$ and $\lambda = 6$: $$\left(\dfrac{1}{3}\right)^{6/3} = \left(\dfrac{1}{3}\right)^{2} = \dfrac{1}{9} \leq \dfrac{1}{9}$$ ✓ The inequality is satisfied with equality. Let us also verify that $\lambda = 5$ does not work for $x = 3$: $$\left(\dfrac{1}{3}\right)^{5/3} = 3^{-5/3} \approx 0.161$$ Since $0.161 > \dfrac{1}{9} \approx 0.111$, the inequality is NOT satisfied. Therefore, the minimum value of $\lambda$ is $\boxed{6}$. **Answer: Option 4**