Using the first given equation: \( \log_a 30 = \log_a (5 \cdot 6) = \log_a 5 + \log_a 6 = A \) (1). Using the second given equation: \( \log_a \left( \frac{5}{3} \right) = \log_a 5 - \log_a 3 = -B \) (2).
Adding equations (1) and (2) yields \( (\log_a 5 + \log_a 6) + (\log_a 5 - \log_a 3) = A - B \), which simplifies to \( 2\log_a 5 + \log_a 6 - \log_a 3 = A - B \) (3). Expanding \( \log_a 6 \) as \( \log_a 2 + \log_a 3 \): \( 2\log_a 5 + (\log_a 2 + \log_a 3) - \log_a 3 = A - B \). This further simplifies to \( 2\log_a 5 + \log_a 2 = A - B \) (4).
From \( \log_a 5 - \log_a 3 = -B \), we get \( \log_a 3 = \log_a 5 + B \) (5). Substituting (5) into the expanded form of (1) (\( A = \log_a 5 + \log_a 2 + \log_a 3 \)): \( A = \log_a 5 + \log_a 2 + (\log_a 5 + B) \). This results in \( A - B = 2\log_a 5 + \log_a 2 \), which is consistent with equation (4).
The change of base formula is \( \log_3 a = \frac{\log_2 a}{\log_2 3} \). We are given \( \log_2 a = \frac{1}{3} \).
We can write \( \log_2 3 = \frac{\log_2 a}{\log_a 3} = \frac{1/3}{\log_a 3} = \frac{1}{3\log_a 3} \). Substituting this back into the expression for \( \log_3 a \): \( \log_3 a = \frac{\frac{1}{3}}{\frac{1}{3\log_a 3}} = \log_a 3 \).
From Step 3, we have \( \log_a 3 = \log_a 5 + B \). From Step 1, \( \log_a 30 = A = \log_a 5 + \log_a 6 \). Combining these results, and noting that \( \log_3 a = \log_a 3 \), we find that \( \log_3 a = \log_a 3 = A + B \).
\[ \boxed{\log_3 a = A + B} \]
If \[ \log_{p^{1/2}} y \times \log_{y^{1/2}} p = 16, \] then find the value of the given expression.