JAMB - Mathematics (2006 - No. 17)
Calculate the logarithm to base 9 of 3\(^{-4}\) x 9\(^2\) x (81)\(^{-1}\)
2
zero
-2
-4
Explanation
\(3^{-4}\times 9^2 \times 81^{-1}\)
=\(log_9 (3^{-4}\times 9^2 \times 81^{-1})\)
= \(log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\)
= \(log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\)
= \(log_9 \frac{1}{81}\)
= \(log_9 \frac{1}{9^2}\)
= \(log_9 9^{-2}\)
= -2\(log_9 9\)
- 2 \(\times 1\)
= -2
=\(log_9 (3^{-4}\times 9^2 \times 81^{-1})\)
= \(log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\)
= \(log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\)
= \(log_9 \frac{1}{81}\)
= \(log_9 \frac{1}{9^2}\)
= \(log_9 9^{-2}\)
= -2\(log_9 9\)
- 2 \(\times 1\)
= -2
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