$${\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + }$$ $${\frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + }$$ $${\frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}}$$ is equal to -
A. 1
B. $$\frac{3}{2}$$
C. 2
D. 3
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr} $$Related Questions on Logarithm
Which of the following statements is not correct?
A. log10 10 = 1
B. log (2 + 3) = log (2 x 3)
C. log10 1 = 0
D. log (1 + 2 + 3) = log 1 + log 2 + log 3
$${{\log \sqrt 8 } \over {\log 8}}$$ is equal to:
A. $$\frac{1}{6}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{1}{8}$$
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