Question

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Answer 1

\[\frac{ 1 }{ 1+logb ^{a}+logb ^{c} }+\frac{ 1 }{ 1+logc ^{a}+logc ^{b} }+\frac{ 1 }{ 1+loga ^{b}+loga^{c} }\]

Answer 2

This is equal to what

Answer 3

1/(1 + (a+c)log b) + 1/(1 + (a+b) log c) + 1/ (1 + (b+c) loga)

This should be the first step, I believe.

Answer 4

It should be multiplied rather than added right
like:-
logb(ac)

Answer 5

I didn't apply that identity here.

Let me give you example from the first fraction :

\(\dfrac{1}{1 + \log b^a + \log b^c }\)

\(\dfrac{1}{1 + a\log b + c\log b}\)
\(\dfrac{1}{1 + (a+c) \log b}\)

Answer 6

SO if you take that up it would become
logb(a+c)
so
logb(ac)=logb(a+c)

Answer 7

No..! See, I have taken log (b) common and (a+c) is not under logarithm condition.

Answer 8

Well, is there any base given in the question?

Answer 9

no so its natural log

Answer 10

Yeah get your point

Answer 11

What would be after that