1. ## Another Log Question

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

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

2. Originally Posted by Winding Function
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 $\frac{k+x}{k+y}$.