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Question
Compute log9 27 – log27 9
Solution
log927 – log279 = log9 33 – log27 32
= 3 log9 3 – 2 log27 3 ......(1)
By change of base rule [logb a = `1/log_"a""b"`]
(1) ⇒ log927 – log279 = `3/log_3 9 - 2/log_3 27`
= `3/log_3 3^2 - 2/log_3 3^3`
= `3/(2log_3 3) - 2/(3log_3 3)`
= `3/2 - 2/3` ....[∵ log3 3 = 1]
= `(9 - 4)/6`
= `5/6`
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