\(log_810\) = X = \(log_8{2 x 5}\) \(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks OR Given \(\log_8 10 = X\), we can express \(\log_8 5\) as follows: \(\log_8 5 = \log_8 \left( \frac{10}{2} \right) = \log_8 10 - \log_8 2\) Using the change of base formula, we find: \(\log_8 2 = \frac{\log 2}{\log 8} = \frac{1}{3}\) Thus, substituting back, we get: \(\log_8 5 = X - \log_8 2 = X - \frac{1}{3}\)