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.