Denote by $\mu(A_t)$ the max spectral radius of $A_t$ which is defined as the maximal cycle geometric mean $$ \mu(A_t) := \max \{ (a_{i_1i_2}a_{i_2i_3}\dots a_{i_ki_1})^{1/k}\}$$ where the maximum is taken over all cycles in the matrix $A$, $k$ is the length of the cycle, and for each cycle the indices $i_1,\ldots,i_k$ are distinct (so that it is really a cycle). Here this means that $$ \mu(e^t) := \mu(A_{e^t}) = \exp\left( \frac{\ell}{k} t\right)$$ for some maximal cycle of length $k$ and the powers along this cycle sum to $\ell$. In particular, the maximum is attained in the same cycle for all $t\geq0$.

If $n$ is the size of the matrix, then it is known for any nonnegative matrix $B$ that $$ \mu(B) \leq \Lambda(B) \leq n \mu(B), $$ (equation (6.10) in Elsner, Johnson, Dias da Silva, The Perron root of a weighted geometric mean of nonnegative matrices. Linear and Multilinear Algebra 24 (1988) 1-13.)

So here this means $$ \exp\left( \frac{\ell}{k} t\right) \leq \Lambda(e^t) \leq n\exp\left( \frac{\ell}{k} t\right)$$ where the $\ell/k$ is actually easy to get from the matrix $A_t$.

Denote by $\mu(A_t)$ the max spectral radius of $A_t$ which is defined as the maximal cycle geometric mean $$ \mu(A_t) := \max \{ (a_{i_1i_2}a_{i_2i_3}\dots a_{i_ki_1})^{1/k}\}$$ where the maximum is taken over all cycles in the matrix $A$, $k$ is the length of the cycle, and for each cycle the indices $i_1,\ldots,i_k$ are distinct (so that it is really a cycle). Here this means that $$ \mu(e^t) := \mu(A_{e^t}) = \exp\left( \frac{\ell}{k} t\right)$$ for some maximal cycle of length $k$ and the powers along this cycle sum to $\ell$. In particular, the maximum is attained in the same cycle for all $t\geq0$, namely one in which sum of the exponents divided by length of the cycle is maximized. For $t<0$ on the other hand we need to minimize the factor $\ell/k$ over all cycles.

If $n$ is the size of the matrix, then it is known for any nonnegative matrix $B$ that $$ \mu(B) \leq \Lambda(B) \leq n \mu(B), $$ (equation (6.10) in Elsner, Johnson, Dias da Silva, The Perron root of a weighted geometric mean of nonnegative matrices. Linear and Multilinear Algebra 24 (1988) 1-13.)

So here this means $$ \exp\left( \frac{\ell}{k} t\right) \leq \Lambda(e^t) \leq n\exp\left( \frac{\ell}{k} t\right),\quad t\geq0$$ where the $\ell/k$ is actually easy to get from the matrix $A_t$. Similarly, but with a different constant (in general) for $t<0$.