\(log_810\) = 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: \(\log_8 5 = X - \log_8 2 = X - \frac{1}{3}\)