Explanation: (C) The given operations suggest the formula for \(a \# b\) is \(a^{b'} - b''\) where \(b'\) is a value related to b. Let's try \(a \# b = a^2 - b\). \(4 \# 2 = 4^2 - 2 = 16 - 2 = 14\). (Matches). \(2 \# 3 = 2^2 - 3 = 4 - 3 = 1\). (Doesn't match 6). Let's try \(a \# b = a^b - a\). \(4 \# 2 = 4^2 - 4 = 16 - 4 = 12\). (Doesn't match 14). Let's try the pattern in the provided explanation: \((a^2)-b\) for the first case: \(4^2 - 2 = 14\). And \((a^3)-b\) for the second case: \(2^3 - 2 = 6\) (the source says 2#3=6 but shows \(2^3-2=6\), implying the general formula is \(a^{b_0} - b_1\)). Since the numbers are small, the simplest pattern is likely the one that works for both: \(4 \# 2 = 14\). \(2 \# 3 = 6\). Let's try \(a \# b = a^b - a - b\). \(4^2 - 4 - 2 = 10\). No. Let's try \(a \# b = (a+b) \cdot b - b\). \( (4+2) \cdot 2 - 2 = 12 - 2 = 10\). No. The logic provided in the source is \(4 \# 2 = 4^2 - 2 = 14\) and \(2 \# 3 = 2^3 - 2 = 6\). The formula appears to be \(a \# b = a^{\text{something}} - 2\). For \(4 \# 2\), power is 2. For \(2 \# 3\), power is 3. It seems the power is \(b\). Let's assume the operator is \(a \# b = a^b - 2\). \(5 \# 2 = 5^2 - 2 = 25 - 2 = 23\). This matches the option. So, the correct option is C.
