Are vector of log terms linearly independent let $\{x_i, 1 \leq i \leq n\}$ and  $\{y_i, 1 \leq i \leq n\}$ be two finite sequences of stricly positive real numbers such that $x_i < x_{i+1}$ and $y_i < y_{i+1}$. Consider the matrix $M$ whose (i,j)th element is 
\begin{eqnarray}
M(i,j) = \log \left(1+ \frac{x_i}{y_j}\right).
\end{eqnarray}
Is M a full rank matrix?
A partial idea i have tried is this. 
Suppose not, then there exist numbers $z_j, 1 \leq j \leq n$ such that not all of them are zero for all and for all $1 \leq i \leq n$ 
\begin{eqnarray}
\sum_{j=1}^{n} \log \left(1+ \frac{x_i}{y_j}\right)z_j =0.
\end{eqnarray}
Consider the function 
\begin{eqnarray}
f(x) = \sum_{j=1}^{n} \log\left(1+ \frac{x}{y_j}\right)z_j =0,
\end{eqnarray}
then $f(x)$ is zero at n distinct values, by mean value theorem the derivative of $f(x)$ is zero at $n-1$ distinct values.
The derivative is 
\begin{eqnarray}
f'(x) = \sum_{j=1}^{n} \frac{ z_j}{1+x + y_j}.
\end{eqnarray}
I wanted to show that the derivative is a rational function and can have atmost $n-1$ zeroes and contadict the result, but for that i needed  the derivative of $f(x)$ to be zero at more than $n-1$ distinct values.
I think i got the proof, One can also observe that the function $f(x)$ is zero at $x=0$. By mean value theorem this will provide the additional point where $f'(x) = 0$, hence completing the above proof.  
 A: Let $(x_i)_{i=1}^n, (y_j)_{j=1}^n$ be as above. We assume that $(\log(1+x_i/y_j))_{1\leq i,j \leq n}$ has not full rank. Hence, the kernel is nontrivial and therefore exists $(z_k)_{k=1}^n\in \mathbb{R}^n$ such that for all $i\in \{1, \dots, n \}$ holds
$$ \sum_{j=1}^n \log\left( 1 + \frac{x_i}{y_j} \right) z_j =0.$$
We define 
$$f(x):= \sum_{j=1}^n \log\left( 1 + \frac{x}{y_j} \right) z_j.$$
We see that $f$ has at least $n+1$ zeros. Namely $x_i$ for $i\in \{1, \dots, n \}$ and zero (they are distinct as $0<x_i <x_{i+1}$). By the mean value theorem we get that $f'$ has at least $n$ zeros. However, we compute
$$ f'(x) = \sum_{j=1}^n  \frac{1}{1+\frac{x}{y_j}}\cdot\frac{1}{y_j}\cdot z_j
= \sum_{j=1}^n \frac{z_j}{x+y_j}
= \frac{1}{\prod_{l=1}^n (x+y_l)} \underbrace{\sum_{j=1}^n z_j \prod_{\substack{k=1\\k\neq j}}^n (x+y_k)}_{=:g(x)}.$$
Note that $f'(x)=0$ iff $g(x)=0$. But $g$ is a polynomial of degree at most $n-1$ and has therefore at most $n-1$ zeros (if $g$ is not the zero polynomial). This contradicts the fact that $f'$ has at least $n$ zeros.
Hence, we are left to show that $g$ is not the zero polynomial. If $g$ was the zero polynomial, we conclude that $f'$ is identically zero. This would imply that $f$ is constant. However, we can write
$$ f(x)= \sum_{j=1}^n \log\left( 1 + \frac{x}{y_j} \right) z_j
= \log\left( \prod_{j=1}^n \left( 1 + \frac{x}{y_j} \right)^{z_j}  \right).$$
This would imply that
$$h(x):= \prod_{j=1}^n \left( 1 + \frac{x}{y_j} \right)^{z_j} $$
is constant. However, one easily computes that $h'(x)>0$ which gives us the desired contradiction.
