10th Class Mathematics Related to Competitive Exam Question Bank Algebra

  • question_answer
    If \[x=\frac{y}{{{(1+a)}^{p}}}\], then p is equal to

    A)  \[\frac{{{\log }_{e}}\left( \frac{y}{x} \right)}{{{\log }_{e}}(1+a)}\]              

    B)         \[\log \left\{ \frac{y}{x(1+a)} \right\}\]

    C)  \[\log \left\{ \frac{y-x}{1+a} \right\}\]  

    D)         \[\frac{\log y}{\log \{x(1+a)\}}\]

    Correct Answer: A

    Solution :

     Since    \[x=\frac{y}{{{(1+a)}^{p}}}\] \[\therefore \]  \[{{(1+a)}^{p}}=\frac{y}{x}\] or            \[p{{\log }_{e}}(1+a)={{\log }_{e}}\frac{y}{x}\] or            \[p=\frac{{{\log }_{e}}\left( \frac{y}{x} \right)}{{{\log }_{e}}(1+a)}\]


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