ABC26GN2075 · Logarithms
Subject: General Aptitude · Chapter: Logarithms · Exam: 2005 · Marks: · Difficulty:
$\left[\frac{1}{\left(\log _{a} b c\right)+1}+\frac{1}{\left(\log _{b} c a\right)+1}+\frac{1}{\left(\log _{c} a b\right)+1}\right]$ is equal to
(a)1
(b)$\frac{3}{2}$
(c)2
(d)3
Answer
Explanation
Given expression $$\begin{aligned} & =\frac{1}{\log _{a} b c+\log _{a} a}+\frac{1}{\log _{b} c a+\log _{b} b}+\frac{1}{\log _{c} a b+\log _{c} c} \\ & =\frac{1}{\log _{a}(a b c)}+\frac{1}{\log _{b}(a b c)}+\frac{1}{\log _{c}(a b c)} \\ & =\log _{abc} a+\log _{abc} b+\log _{abc} c \\ & =\log _{abc}(a b c)=1 \end{aligned}$$
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