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logaN. logbN+logcN. logbN+logaN. logcNis...

`log_aN. log_bN+log_cN. log_bN+log_aN. log_cN`is equals to

A

`(log_(a)N. log_(b)N. log_(c )N)/(log_(abc)N)`

B

`(log_(abc)N)/(log_(a)N. log_(b)N. log_(c )N)`

C

`(log_(N)abc)/(log_(N)a. log_(N)b. log_(N)c)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`log_(a)N.log_(b)N+log_(b)N.log_(c )N+log_(c )N. log_(a)N`
`=(1)/(log_(N)a alog_(N)b)+(1)/(log_(N)b log_(N)c)+(1)/(log_(N)c log_(N)a)`
`=(log_(N)a+log_(N)b+log_(N)c)/((log_(N)a)(log_(N)b)(log_(N)c))`
`=(log_(N)abc)/((1)/(log_(a)N)(1)/(log_(b)N)(1)/(log_(c )N))`
`=(log_(a)N.log_(b)N.log_(c )N)/(log_(abc)N)`
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