Question:

Step 1: Evaluate option (a). Let $x = \sqrt{7\sqrt{7\sqrt{7\cdots}}}$. Then $x = 7^{1/2} \cdot 7^{1/4} \cdot 7^{1/8} \cdots = 7^{1/2 + 1/4 + 1/8 + \cdots} = 7^1 = 7$. So $\log_5 7 > \log_5 5 = 1$. Option (a) is correct. Step 2: Evaluate option (b). Note $\sqrt{7} - \sqrt{6} < 1$ (since $\sqrt{7} \approx 2.646$, $\sqrt{6} \approx 2.449$, difference $\approx 0.197$). Also $\sqrt{3} - \sqrt{2} < 1$. For a base $0 < b < 1$, $\log_b x < 1$ iff $x > b$. We need $\sqrt{3}-\sqrt{2} > \sqrt{7}-\sqrt{6}$. Since $\sqrt{3}-\sqrt{2} \approx 0.318$ and $\sqrt{7}-\sqrt{6} \approx 0.197$, yes. So $\log_{(\sqrt{7}-\sqrt{6})}(\sqrt{3}-\sqrt{2}) < 1$. Option (b) is correct. Step 3: Evaluate option (c). $\log_3 10 = \frac{\ln 10}{\ln 3} \approx \frac{2.303}{1.099} \approx 2.096$. $\log_{10} 70 = \log_{10}(7 \times 10) = 1 + \log_{10} 7 \approx 1 + 0.845 = 1.845$. Since $2.096 > 1.845$, option (c) is correct. Step 4: Evaluate option (d). $\log_3(3+\sqrt{2}) > \log_3 3 = 1$ since $3+\sqrt{2} > 3$. For $\log_2(2-\sqrt{2})$: $2 - \sqrt{2} \approx 0.586 < 1$, so $\log_2(2-\sqrt{2}) < 0$. Since $\log_3(3+\sqrt{2}) > 1 > 0 > \log_2(2-\sqrt{2})$, option (d) is correct. Step 5: All four options are correct.