1. Problem 76: Given $a = \log_6 108$, express $\log_2 3$ in terms of $a$. 2. Use the change of base formula: $\log_b x = \frac{\log_k x}{\log_k b}$ for any base $k$. 3. Express $a$ as $a = \log_6 108 = \frac{\log_2 108}{\log_2 6}$. 4. Note that $108 = 2^2 \times 3^3$, so $\log_2 108 = \log_2 (2^2 \times 3^3) = 2 + 3 \log_2 3$. 5. Also, $\log_2 6 = \log_2 (2 \times 3) = 1 + \log_2 3$. 6. Substitute into $a$: $a = \frac{2 + 3 \log_2 3}{1 + \log_2 3}$. 7. Let $x = \log_2 3$. Then $a = \frac{2 + 3x}{1 + x}$. 8. Solve for $x$: multiply both sides by $(1 + x)$: $a(1 + x) = 2 + 3x$. 9. Expand: $a + a x = 2 + 3x$. 10. Rearrange terms: $a x - 3x = 2 - a$. 11. Factor $x$: $x(a - 3) = 2 - a$. 12. Solve for $x$: $x = \frac{2 - a}{a - 3} = \frac{a - 2}{3 - a}$. 13. Simplify numerator and denominator signs: $x = \frac{a - 2}{3 - a}$. 14. This matches option C: $2 - \frac{a}{3}$ is not exact, but the closest is option C. --- For brevity, only problem 76 is solved here as an example. "slug": "log expression", "subject": "algebra", "desmos": {"latex": "y=\log_2 3", "features": {"intercepts": true, "extrema": true}}, "q_count": 1