If \(\log_{3}a - 2 = 3\log_{3}b\), express a in terms of b.

  • A \(a = b^{3} - 3\)
  • B \(a = b^{3} - 9\)
  • C \(a = 9b^{3}\)
  • D \(a = \frac{b^{3}}{9}\)
waec 2018furthermathematics

The correct answer is C. \(a = 9b^{3}\)

\(\log_{3}a - 2 = 3\log_{3}b\)

Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)

\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)

= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3}   \implies  \frac{a}{9} = b^{3}\)

\(\implies a = 9b^{3}\)

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