Question 1034314
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Please help me solve this question
Prove the identity ; log N base a . Log N base b + log N base b . Log N base c + log N base c . Log N base a = log N base a . Log N base b . Log N base c/ log N base abc
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Check that the left side is equal to

{{{1/(log(N,(a))*log(N,(b)))}}} + {{{1/(log(N,(b))*log(N,(c)))}}} + {{{1/(log(N,(a))*log(N,(c)))}}}.


For it, use the identity {{{log(x,(N))}}} = {{{1/log(N,(x))}}}, which is universal for logarithms.


Write this sum with the common denominator, which is

{{{log(N,(a))*log(N,(b))*log(N,(c))}}}.

Then the numerator will be 

{{{log(N,(a))+log(N,(b))+log(N,(c))}}} = {{{log(N,(abc))}}} = {{{1/log((abc),N))}}}.

Exactly as the right side is.


Proved.
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