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log (log(ab) a +(1)/(log(b)ab) is (where...

`log (log_(ab) a +(1)/(log_(b)ab)` is (where `ab ne 1)`

A

0

B

1

C

`log_(a)ab`

D

`log_(b)ab`

Text Solution

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The correct Answer is:
B
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(1)/(log_(a)(ab))+(1)/(log_(b)(ab))=1

(1)/(log_(a)(ab)+(1)/(log_(b)(ab)=1))

Knowledge Check

  • If log_(a)(ab)=x , then log_(b) (ab) is equal to

    A
    `(1)/(x)`
    B
    `(x)/(1+x)`
    C
    `(x)/(x-1)`
    D
    `(x)/(1-x)`

  • (1/(log_(a)bc+1) + 1/(log_(b) ac +1) + 1/(log_(c)ab+1)+1) is equal to:

    A
    1
    B
    2
    C
    0
    D
    abc

  • (1)/(1+log_(a) bc) + (1)/(1+log_(b)ca) + (1)/(1+ log_(c ) ab) is equal to

    A
    0
    B
    3
    C
    2
    D
    1

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    If log_(a)(ab)=x then log_(b)(ab) is equals to

    1/(log_(ab)abc) + 1/(log_(bc) abc) + 1/(log_(ca)abc) is equal to:

    If a, b, c are positive real numbers, then (1)/("log"_(ab)abc) + (1)/("log"_(bc)abc) + (1)/("log"_(ca)abc) =

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