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Question
Prove that loga2 a + logb2 b + logc2 c = `1/8`
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Solution
By change of base rule logb a = `1/log_"a""b"`
loga2 a × logb2 b × logc2 c = `1/log_"a""a"^2 xx 1/log_"b""b"^2 xx 1/log_"c""c"^2`
= `1/(2log_"a""a") xx 1/(2log_"b""b") xx 1/(2log_"c""c")`
= `1/2 xx 1/2 xx 1/2`
= `1/2^3`
= `1/8`
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