Calculate the logarithm to base 9 of \(3^{-4} \times 9^2 \times (81)^{-1}\)

  • A 2
  • B zero
  • C -2
  • D -4
jamb 2006mathematics

The correct answer is C. -2

\(3^{-4}\times 9^2 \times 81^{-1}\)

\(=log_9 (3^{-4}\times 9^2 \times 81^{-1})\)

\(=log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\)

\(=log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\)

\(=log_9 \frac{1}{81}\)

\(=log_9 \frac{1}{9^2}\)

\(=log_9 9^{-2}\)

\(=-2log_9 9=-2 \times 1=-2\)

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