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Thread: Another Log Question

  1. #1
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    Another Log Question

    Let $\displaystyle x$, $\displaystyle y$, $\displaystyle a$, $\displaystyle k$ be positive real numbers such that $\displaystyle x > y > a > 1$. Prove that

    $\displaystyle 1+\log_{a} \frac{x}{y} > \log_{k+y}(k+x)$.
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  2. #2
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    Quote Originally Posted by Winding Function View Post
    Let $\displaystyle x$, $\displaystyle y$, $\displaystyle a$, $\displaystyle k$ be positive real numbers such that $\displaystyle x > y > a > 1$. Prove that

    $\displaystyle 1+\log_{a} \frac{x}{y} > \log_{k+y}(k+x)$.
    $\displaystyle 1+\log_{a} \frac{x}{y} >1+\log_{k+y}(\frac{k+x}{k+y})$

    $\displaystyle \log_{a} \frac{x}{y} >\log_{k+y}(\frac{k+x}{k+y})$

    Note that:

    $\displaystyle k+y>a>1$

    and:

    $\displaystyle \frac{x}{y}>\frac{k+x}{k+y}$

    So, for $\displaystyle a$, which is smaller than $\displaystyle k+y$, to reach $\displaystyle \frac{x}{y}$, which is a higher number than $\displaystyle \frac{k+x}{k+y}$, it must be raised to a higher power than $\displaystyle k+y$ must be raised to reach $\displaystyle \frac{k+x}{k+y}$.
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