<p>Given \(\log_3 a = p = \log_b c\) and \(\log_b 9 = \frac{2}{p^2}\), if \(\log_9\!\left(\frac{a^4 b^3}{c}\right) = \alpha p^3 + \beta p^2 + \gamma p + \delta\), then \((\alpha + \beta + \gamma + \delta)\) equals</p>
Step-by-Step Solution
Key Concept: Express everything in base 9 using the three given conditions. From log_3 a = p we get a = 3^p. From log_b 9 = 2/p^2, we get log_9 b = p^2/2. Since log_b c = p, we have log_9 c = p^3/2. Therefore log_9((a^4 b^3)/c) = ...
Notice that the cleanest route is to simplify the structure before computing. A clever move here is to translate the logarithmic statement into a friendlier algebraic form. Express everything in base 9 using the three given conditions. From log_3 a = p we get a = 3^p. From log_b 9 = 2/p^2, we get log_9 b = p^2/2. Since log_b c = p, we have log_9 c = p^3/2. Therefore log_9((a^4 b^3)/c) = 4(p/2) + 3(p^2/2) - p^3/2, giving a polynomial whose coefficients sum to 3. Trap: Translate each condition into base-9 logs first; direct substitution in mixed bases is messy. Now, we invoke the power of the relevant logarithmic identity, simplify carefully, and finally verify the domain so that no extraneous answer survives.
Correct Answer: C