52. Condense the expression to the logarithm of a single quantity. \( \frac{1}{2}\left[\log _{7}(x+8)+2 \log _{7}(x-8)\right]-16 \log _{7} x \) a. \( \log _{7} \frac{x^{16} \sqrt{x+8}}{(x-8)} \) b. \( \log _{7} \frac{x^{2} \sqrt{x-8}}{(x+8)} \) c. \( \log _{7} \frac{(x-8) \sqrt{x+8}}{x^{16}} \) d. \( \log _{7} \frac{(x+8) \sqrt{x-8}}{x^{2}} \)

To condense the given expression: Firstly, apply the properties of logarithms. The term \( \frac{1}{2}\left[\log _{7}(x+8)+2 \log _{7}(x-8)\right] \) can be re-written as \( \frac{1}{2}\log_{7}((x+8)(x-8)^2) \) because \( 2 \log_{7}(x-8) \) becomes \( \log_{7}((x-8)^2) \). Then, applying the power rule gives us a factor of \( \frac{1}{2} \). This yields: \[ \frac{1}{2} \log_{7} \left( (x+8)(x-8)^2 \right) = \log_{7} \left( \sqrt{(x+8)(x-8)^2} \right) = \log_{7} \left( \frac{\sqrt{(x+8)(x-8)^2}}{x^{16}} \right) \] Finally, minus \( 16 \log_{7} x \) means we combine with it, leading to: \[ \log_{7} \left( \frac{\sqrt{(x+8)(x-8)^2}}{x^{16}} \right) \] This simplifies down to: \[ \log_{7} \frac{(x-8) \sqrt{x+8}}{x^{16}} \] So the answer is: c. \( \log_{7} \frac{(x-8) \sqrt{x+8}}{x^{16}} \)