9th Class Mathematics Related to Competitive Exam Question Bank Logarithms

  • question_answer
    \[log\left( \frac{1}{2} \right)+log\left( \frac{2}{3} \right)+log\left( \frac{3}{4} \right)+......+log\left( \frac{999}{10000} \right)\text{= }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ }\text{.}\]

    A)  \[-3\]      

    B)  \[-1\]            

    C)  0                                

    D)  2

    Correct Answer: A

    Solution :

    (a): \[log\left( \frac{1}{2} \right)+log\left( \frac{2}{3} \right)+......+log\left( \frac{999}{1000} \right)\] \[=log1-log2+log2-log3+....+log999-log1000\] \[=log1-log1000=0-log{{10}^{3}}\] \[=-\,3log10=-\,3.\]                    


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