<p>If \(2^{x+y} = 6^y\) and \(3^{x-1} = 2^{y+1}\), then the value of \(\dfrac{\log 3 - \log 2}{x-y}\) is</p>
Step-by-Step Solution
Key Concept: Take common logarithms of both equations and solve for x - y. From the first equation, x log 2 = y log 3. From the second, (x - 1)log 3 = (y + 1)log 2. Eliminating gives x - y = 1, so the required value is log 3 - log...
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. Take common logarithms of both equations and solve for x - y. From the first equation, x log 2 = y log 3. From the second, (x - 1)log 3 = (y + 1)log 2. Eliminating gives x - y = 1, so the required value is log 3 - log 2 = log(3/2). Trap: Once x - y = 1 is found, both option forms become the same numerical value. 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, D