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Question: $\frac{1}{1 + log_{b} a + log_{b} c} + \frac{1}{1 + log_{c} a + log_{c} b} + \frac{1}{1 + log_{a} b ...

11+logba+logbc+11+logca+logcb+11+logab+logac\frac{1}{1 + log_{b} a + log_{b} c} + \frac{1}{1 + log_{c} a + log_{c} b} + \frac{1}{1 + log_{a} b + log_{a} c}

A

abc

B

1abc\frac{1}{abc}

C

0

Answer

1

Explanation

Solution

Using the properties 1+logxy+logxz=logx(xyz)1 + log_x y + log_x z = log_x (xyz) and 1logxy=logyx\frac{1}{log_x y} = log_y x, the expression simplifies to logabcb+logabcc+logabca=logabc(abc)=1log_{abc} b + log_{abc} c + log_{abc} a = log_{abc} (abc) = 1.