The sentence $1+1=2$ is not logically equivalent with $99 +1 =100$.
To see this note the way logical equivalence is defined in model theory. In first-order logic two sentences $\varphi, \psi$ of some first-order language $L$ are logically equivalent iff $M$ models $\varphi$ exactly if $M$ models $\psi$, for every $L$-model $M$. The definition quantifies over every model for $L$, even if $L$ is the language of number theory. And there are no logical constraints on how to interpret non-logical constants like numerals or the addition symbol (besides assigning them objects of the right type).
To be more specific take $L$ to be the sublanguage of first-order Peano Arithmetic with the signature $\{+,~ ', ~0 \}$. For any non-zero natural number $n$ we may define the constant $\underline{n} := 0' \ldots'$ ($n$ bars). Let $M = (\mathbb{N}, +_{M},~ '_{M},~ 0_{M})$, where $'_{M}$ is the successor function, $0_{M}$ is the number $0$, but $+_{M} = \{((1,1), 2)\}$. Then it's immediate that $M$ models the $L$-sentence $\underline{1}+\underline{1} = \underline{2}$, but not $\underline{99} +\underline{1} = \underline{100}$.
Of course it is possible to restrict the class of $L$-models to those models which are isomorphic to the $L$-reduct of the standard model of arithmetic (setting aside non-standard models for the moment). Relative to this restricted class your example sentences indeed are always either both true or both false. But that does not amount to logical equivalence proper.
