Are translations of logarithms linearly independent? I think I proved the following but I am not sure. I will write my answer at the bottom
Is the set of logarithms $ \lbrace\ln (t + a_i)\rbrace_{i=1}^N $ with $t,a_i>0$, and all $a_i$ different linearly independent?
$N$ is a given integer. 
Edit: 
My solution is the following:
The condition for linear dependence is that the function
\begin{equation}
f(t) = \sum_{i=1}^N c_i \ln (t + a_i) \equiv 0 
\end{equation}
for some $c_i$ not all zero. 
The function can be rewritten into
\begin{equation}
f(t) = \ln \left(\prod_{i=1}^N (t + a_i)^{c_i} \right) \equiv 0
\end{equation}
which implies that the function
\begin{equation}
g(t) = \prod_{i=1}^N (t + a_i)^{c_i} \equiv 1
\end{equation}
note that $g(t)$ can never be zero due to the hypothesis on $t$ and $a_i$.
Taking derivatives with respect to time we obtain
\begin{equation}
g'(t) = \left( \sum_{i=1}^{N}\frac{c_i}{t+a_i}\right)\prod_{i=1}^N (t+a_i) ^{c_i} = \sum_{i=1}^{N}\frac{{c}_{i}}{t+{a}_{i}} g(t) \equiv 0
\end{equation}
Since $g(t)$ is never zero (and always one) we can remove it and obtain the equality
\begin{equation}
\sum_{i=0}^{N}\frac{{c}_{i}}{t+{a}_{i}} \equiv 0
\end{equation}
Note that this states that translations of negative powers (of degree -1) are also linearly dependent.
Using the idea in this answer we can remove the denominators obtaining 
\begin{equation}
\sum_{i=1}^N {c_i}\prod_{j\neq i}(t + a_j) \equiv 0.
\end{equation}
The left-hand side of that equation defines a polynomial of degree $N-1$ in the variable $t$. Evaluating it at $N$ different values of $t$ produces a contradiction, since a polynomial of degree $N-1$ cannot have $N$ roots. Hence all $c_i$ must be zero.
Therefore 
a) Translations of negative powers (of degree -1) are linearly independent.
b) $g(t) \not\equiv 1$.
c) The given logarithms are linearly independent.
 A: They are for $N=2$. I will show that if they are linearly dependent
then $a_1 = a_2$ and $b_1 = -b_2$.
Exponentiating, and making all indexed variables distinct,
this is
$1 = (t+a)^b (t+c)^d$.
Looking at large $t$, this implies that $b$ and $d$ have opposite signs.
Let $b$ be the positive one, and let $D = -d$.
This becomes 
$(t+a)^b = (t+c)^D$
with $b$ and $D$ positive.
Taking the $b$-th root (or looking at large $t$),
we find that $b = D$ and, then, 
$a = c$.
A: Assume for $a_1, \dotsc, a_n$ all different there are $c_1, \dotsc, c_n$, not all zero, so that:
$\begin{align*}
  \sum_{1 \le k \le n} c_k \ln(t + a_k)
    &= 0 \\
  \ln \prod_{1 \le k \le n} (t + a_k)^{c_k}
    &= 0
\end{align*}$
Differentiate:
$\begin{align*}
  \sum_{1 \le k \le n} \frac{c_k}{t + a_k}
    &= 0
\end{align*}$
This is continuous except at $t = -a_1, \dotsc, -a_n$. Without loss of generality, take $c_1 \ne 0$, for $t \to -a_1^+$ the limit is $\infty$, thus for $t$ near enough $-a_1$ the sum is larger than 1, unless $c_1 = 0$, contradiction.
