Write the expression as a single logarithm. \[ 3 \log _{8}(z-3)-5 \log _{8}(z+6) \] log ( \( \square \),

We begin with the expression \[ 3 \log_{8}(z-3) - 5 \log_{8}(z+6). \] Using the logarithm rule \( a\log_b c = \log_b(c^a) \), we can rewrite each term: \[ 3 \log_{8}(z-3) = \log_{8}((z-3)^3), \] \[ 5 \log_{8}(z+6) = \log_{8}((z+6)^5). \] Then, using the difference property of logarithms \[ \log_b A - \log_b B = \log_b\left(\frac{A}{B}\right), \] we combine the two logarithms: \[ \log_{8}((z-3)^3) - \log_{8}((z+6)^5) = \log_{8}\left(\frac{(z-3)^3}{(z+6)^5}\right). \] Thus, the original expression as a single logarithm is: \[ \log_{8}\left(\frac{(z-3)^3}{(z+6)^5}\right). \]