(c) Evaluate the following expression for A without using tables or \( A=\frac{4 \log 9-2 \log 27+\log 243}{7 \log 3} \)

To evaluate the expression for \( A \), we can start by simplifying the logarithms. First, we know that: - \( \log 9 = \log(3^2) = 2\log 3 \) - \( \log 27 = \log(3^3) = 3\log 3 \) - \( \log 243 = \log(3^5) = 5\log 3 \) Now substitute these into the expression for \( A \): \[ A = \frac{4(2\log 3) - 2(3\log 3) + 5\log 3}{7 \log 3} \] This simplifies to: \[ A = \frac{8\log 3 - 6\log 3 + 5\log 3}{7 \log 3} \] Combining the terms in the numerator gives: \[ A = \frac{(8 - 6 + 5)\log 3}{7 \log 3} = \frac{7 \log 3}{7 \log 3} \] Since the \( \log 3 \) terms cancel out, we find: \[ A = 1 \] So, the final answer is \( A = 1 \).