If log\(_810\) = X, evaluate log\(_85\) in terms of X.
- \(\frac{1}{2}\)X
- X-\(\frac{1}{4}\)
-
X-\(\frac{1}{3}\)
- X-\(\frac{1}{2}\)
The correct answer is: C
Explanation
\(log_810\) = 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