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If log\(_810\) = X, evaluate log\(_85\) in terms of X.

Mathematics
JAMB 1999

If log\(_810\) = X, evaluate log\(_85\) in terms of X.

  • A. \(\frac{1}{2}\)X
  • B. X-\(\frac{1}{4}\)
  • C. X-\(\frac{1}{3}\)
  • D. X-\(\frac{1}{2}\)
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Correct Answer: Option 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


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