Question

Prove this identity.

Show appropriate calculations.

Answer 1

Hopefully, this identity will not irritate anyone.

Answer 2

LHS =( loga N ).( logb N) +( logb N ).(logc N) + (logc N).( loga N)

we know,
log{ a base b } = 1/log{ b base a }
use this concept here,

= 1/( logN a )( logN b ) + 1/( logN b)(logN c) + 1/( logN c )( logN a )

let (logN a ) = x
(logN b ) = y
( logN c) = z

so,
= 1/xy + 1/yz + 1/zx

= ( x + y + z)/xyz

now put x, y and z value

= { ( logN a ) + ( logN b) + ( logN c ) }/{(lonN a ).( logN b).(logN c) }

base is same of denominator so,

= {logN ( a × b × c)}/ {(logN a ).( logN b).(logN c) }

= 1/(logabc N) /1/ {(loga N ).( logb N).(logc N) }

=( loga N).( logb N).( logc N)/(logabc N ) =RHS