Results 1 to 2 of 2

Math Help - Another Log Question

  1. #1
    Banned
    Joined
    Sep 2008
    Posts
    47

    Another Log Question

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

    1+\log_{a} \frac{x}{y} > \log_{k+y}(k+x).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by Winding Function View Post
    Let x, y, a, k be positive real numbers such that x > y > a > 1. Prove that

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

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

    Note that:

    k+y>a>1

    and:

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

    So, for a, which is smaller than k+y, to reach \frac{x}{y}, which is a higher number than \frac{k+x}{k+y}, it must be raised to a higher power than k+y must be raised to reach \frac{k+x}{k+y}.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum