8th Class Mathematics Related to Competitive Exam Question Bank Logarithms

  • question_answer
    \[\left[ \frac{1}{\left( {{\log }_{x}}yz \right)+1}+\frac{1}{\left( {{\log }_{y}}zx \right)+1}+\frac{1}{\left( {{\log }_{z}}xz \right)+1} \right]\]is equal to:

    A)  1  

    B)  \[\frac{3}{2}\]               

    C)  2                                

    D)  3

    Correct Answer: A

    Solution :

    (a): Given expression = \[\frac{1}{{{\log }_{x}}yz+{{\log }_{x}}x}+\frac{1}{{{\log }_{y}}zx+{{\log }_{y}}y}+\frac{1}{{{\log }_{z}}xz+{{\log }_{z}}z}\]        \[=\frac{1}{{{\log }_{x}}(xyz)}+\frac{1}{{{\log }_{y}}(xyz)}+\frac{1}{{{\log }_{z}}(xyz)}=\] \[={{\log }_{xyz}}x+{{\log }_{xyz}}y+{{\log }_{xyz}}z\] \[={{\log }_{xyz}}(xyz)=1\].


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