1. **State the problem:** Solve the equation $$\sqrt{7} \cdot \frac{\log n}{\log 2} - \frac{\log 2n}{\log 2} = 1$$ for $n$. 2. **Recall logarithm properties:** - $\log(ab) = \log a + \log b$ - $\frac{\log a}{\log b} = \log_b a$ (change of base formula) 3. **Rewrite the equation using these properties:** $$\sqrt{7} \cdot \log_2 n - \log_2 (2n) = 1$$ 4. **Expand $\log_2 (2n)$:** $$\log_2 2 + \log_2 n = 1 + \log_2 n$$ 5. **Substitute back:** $$\sqrt{7} \cdot \log_2 n - (1 + \log_2 n) = 1$$ 6. **Distribute and group like terms:** $$\sqrt{7} \log_2 n - 1 - \log_2 n = 1$$ 7. **Combine $\log_2 n$ terms:** $$(\sqrt{7} - 1) \log_2 n - 1 = 1$$ 8. **Isolate $\log_2 n$:** $$(\sqrt{7} - 1) \log_2 n = 2$$ 9. **Solve for $\log_2 n$:** $$\log_2 n = \frac{2}{\sqrt{7} - 1}$$ 10. **Rationalize the denominator:** $$\log_2 n = \frac{2}{\sqrt{7} - 1} \cdot \frac{\sqrt{7} + 1}{\sqrt{7} + 1} = \frac{2(\sqrt{7} + 1)}{7 - 1} = \frac{2(\sqrt{7} + 1)}{6} = \frac{\sqrt{7} + 1}{3}$$ 11. **Rewrite in exponential form to solve for $n$:** $$n = 2^{\frac{\sqrt{7} + 1}{3}}$$ **Final answer:** $$\boxed{n = 2^{\frac{\sqrt{7} + 1}{3}}}$$