[1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - [1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - \eqalign{ & \left[ {\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + \frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + \frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}} \right] \cr & {\text{is}}\,\,{\text{equal}}\,{\text{to - }} \cr} " /> [1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - [1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - \eqalign{ & \left[ {\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + \frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + \frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}} \right] \cr & {\text{is}}\,\,{\text{equal}}\,{\text{to - }} \cr} " /> [1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - [1(logabc)+1+1(logbca)+1+1(logcab)+1]isequalto - \eqalign{ & \left[ {\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + \frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + \frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}} \right] \cr & {\text{is}}\,\,{\text{equal}}\,{\text{to - }} \cr} " />

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$\begin{aligned} [\frac{1}{(\log_{a} b c) + 1} + \frac{1}{(\log_{b} c a) + 1} + \frac{1}{(\log_{c} a b) + 1}] \\ is equal to - \end{aligned}$ $\eqalign{ & \left[ {\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + \frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + \frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}} \right] \cr & {\text{is}}\,\,{\text{equal}}\,{\text{to - }} \cr}$

11
2³/₂
32
43

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given Expression \\ = \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_{a b c} a + \log_{a b c} b + \log_{a b c} c \\ = \log_{a b c} (a b c) \\ = 1 \end{aligned}

$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr}$

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", "text": "

\begin{aligned} [\frac{1}{(\log_{a} b c) + 1} + \frac{1}{(\log_{b} c a) + 1} + \frac{1}{(\log_{c} a b) + 1}] \\ is equal to - \end{aligned}

$\eqalign{ & \left[ {\frac{1}{{\left( {{{\log }_a}bc} \right) + 1}} + \frac{1}{{\left( {{{\log }_b}ca} \right) + 1}} + \frac{1}{{\left( {{{\log }_c}ab} \right) + 1}}} \right] \cr & {\text{is}}\,\,{\text{equal}}\,{\text{to - }} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given Expression \\ = \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_{a b c} a + \log_{a b c} b + \log_{a b c} c \\ = \log_{a b c} (a b c) \\ = 1 \end{aligned}

$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given Expression \\ = \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_{a b c} a + \log_{a b c} b + \log_{a b c} c \\ = \log_{a b c} (a b c) \\ = 1 \end{aligned}

$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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