<p>The equation \(\log_{x^2} 16 + \log_{2x} 64 = 3\) has</p>
Step-by-Step Solution
Key Concept: Convert both logarithms to base 2 by putting y = log_2 x. The equation simplifies to 4/(2y) + 6/(1 + y) = 3, leading to y = 1 or y = -4/3. Hence x = 2 or 2^(-4/3). So there are two real solutions, exactly one integral...
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. Convert both logarithms to base 2 by putting y = log_2 x. The equation simplifies to 4/(2y) + 6/(1 + y) = 3, leading to y = 1 or y = -4/3. Hence x = 2 or 2^(-4/3). So there are two real solutions, exactly one integral solution, one irrational solution, and no prime solution. Trap: Remember 2x is the base of the second logarithm, so x must be positive and x != 1, 1/2. 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: A, B, C, D