(a) \( \log _{8} 7+\log _{8} 5=\log _{8} \square \) (b) \( \log _{7} \square-\log _{7} 2=\log _{7} \frac{9}{2} \) (c) \( \log _{8} \frac{1}{16}=-4 \log _{8} \square \)

Let's tackle each of these logarithmic equations step-by-step! (a) By using the property of logarithms that states \( \log_b m + \log_b n = \log_b (mn) \), we have: \[ \log_{8} 7 + \log_{8} 5 = \log_{8} (7 \cdot 5) = \log_{8} 35 \] So, the answer is \( \log_{8} 35 \). (b) Here we can utilize the property \( \log_b m - \log_b n = \log_b \left(\frac{m}{n}\right) \). Thus, we can rewrite the equation as: \[ \log_{7} \square - \log_{7} 2 = \log_{7} \left(\frac{\square}{2}\right) = \log_{7} \frac{9}{2} \] From this, we deduce: \[ \frac{\square}{2} = \frac{9}{2} \implies \square = 9 \] Hence, the answer is 9. (c) For this equation, we start with the known logarithmic identity that states \( \log_b (m^k) = k \cdot \log_b m \). Therefore: \[ \log_{8} \frac{1}{16} = \log_{8} (16^{-1}) = -\log_{8} 16 \] Next, we can express 16 as \( 8^{\frac{4}{3}} \), so: \[ -\log_{8} 16 = -\frac{4}{3} \implies \log_{8} \frac{1}{16} = -\frac{4}{3} \] Substituting this into the equation provides: \[ -\frac{4}{3} = -4 \log_{8} \square \] Dividing both sides by -4 gives: \[ \log_{8} \square = \frac{1}{3} \] Converting back from logarithmic form gives us: \[ \square = 8^{\frac{1}{3}} = 2 \] Thus, the answer is 2. To summarize: (a) \( \log_{8} 35 \) (b) 9 (c) 2