<p>If \(\log_2(x^2 + 1) + \log_{13}(x^2 + 1) = \log_2(x^2 + 1)\log_{13}(x^2 + 1)\), where \(x \ne 0\), then \(\log_7(x^2 + 24)\) equals</p>
Step-by-Step Solution
Key Concept: Let y = x^2 + 1 and transform the equation into 1/a + 1/b = 1 form. Set A = log_2 y and B = log_13 y. Then A + B = AB. Dividing by AB gives 1/A + 1/B = 1, i.e. log_y 2 + log_y 13 = 1, so log_y 26 = 1 and y = 26. Hence...
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. Let y = x^2 + 1 and transform the equation into 1/a + 1/b = 1 form. Set A = log_2 y and B = log_13 y. Then A + B = AB. Dividing by AB gives 1/A + 1/B = 1, i.e. log_y 2 + log_y 13 = 1, so log_y 26 = 1 and y = 26. Hence x^2 + 24 = 49 and log_7 49 = 2. Trap: Work with y = x^2 + 1 first; solving directly in x is unnecessary. 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: B