1. ## 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

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

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