Prove that loga2 a + logb2 b + logc2 c = 18 - Mathematics

Prove that log_a² a + log_b² b + log_c² c = `1/8`

Sum

By change of base rule log_b a = `1/log_"a""b"`

log_a² a × log_b² b × log_c² 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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Chapter 2: Basic Algebra - Exercise 2.12 [Page 80]

Chapter 2 Basic Algebra
Exercise 2.12 | Q 8 | Page 80

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