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Mathematics

Evaluate :

1loga bc+1+1logb ca+1+1logc ab+1\dfrac{1}{\text{log}{a} \space bc + 1} + \dfrac{1}{\text{log}{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1}

Logarithms

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Answer

Evaluating,

1loga bc+1+1logb ca+1+1logc ab+11log bclog a+1+1log calog b+1+1log ablog c+11log bc + log alog a+1log ca + log blog b+1log ab + log clog clog alog bc + log a+log blog ca + log b+log clog ab + log clog alog b + log c + log a+log blog c + log a + log b+log clog a + log b + log clog a + log b + log clog a + log b + log c1.\Rightarrow \dfrac{1}{\text{log}{a} \space bc + 1} + \dfrac{1}{\text{log}{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc}}{\text{log a}} + 1} + \dfrac{1}{\dfrac{\text{log ca}}{\text{log b}} + 1} + \dfrac{1}{\dfrac{\text{log ab}}{\text{log c}} + 1} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log bc + log a}}{\text{log a}}} + \dfrac{1}{\dfrac{\text{log ca + log b}}{\text{log b}}} + \dfrac{1}{\dfrac{\text{log ab + log c}}{\text{log c}}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log bc + log a}} + \dfrac{\text{log b}}{\text{log ca + log b}} + \dfrac{\text{log c}}{\text{log ab + log c}} \\[1em] \Rightarrow \dfrac{\text{log a}}{\text{log b + log c + log a}} + \dfrac{\text{log b}}{\text{log c + log a + log b}} + \dfrac{\text{log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow \dfrac{\text{log a + log b + log c}}{\text{log a + log b + log c}} \\[1em] \Rightarrow 1.

Hence, 1loga bc+1+1logb ca+1+1logc ab+1\dfrac{1}{\text{log}{a} \space bc + 1} + \dfrac{1}{\text{log}{b} \space ca + 1} + \dfrac{1}{\text{log}_{c} \space ab + 1} = 1.

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