The answer to both questions is negative. Every countable model $\def\sM{\mathfrak M}\sM$ of $\mathit{VTC}^0$ (a very weak fragment of $I\Delta_0+\Omega_1$) is an exponential integer part of a real-closed exponential field by [1,2], hence if this property has any first-order consequences besides $\mathit{IOpen}$ at all, they are included in $\mathit{VTC}^0$.

The basic idea is to take for $K$ the completion $\mathbf R^\sM$ of the fraction field $\mathbf Q^\sM$ of $\sM$. This is a real-closed field as $\sM\models\mathit{IOpen}$, and with some effort, one can show that the usual exponential function $2^n\colon\mathbf L^\sM\to\sM$ (where its domain $\mathbf L^\sM$ is the set of logarithmic natural numbers of $\sM$) extends to an isomorphism $2^x\colon(\mathbf R_{\mathbf L}^\sM,+,{<})\simeq(\mathbf R_{>0}^\sM,\cdot,{<})$, where $\mathbf R_{\mathbf L}^\sM=\{x\in\mathbf R^\sM:\exists n\in\mathbf L^\sM\,|x|\le n\}$ is the set of logarithmically bounded reals of $\sM$. Moreover, if $\sM$ is nonstandard, then $(\mathbf Q^\sM,\mathbf Z^\sM,\mathbf Q_{\mathbf L}^\sM,+,{<})$ is recursively saturated, where $\mathbf Z^\sM$ is the extension of $\sM$ with negative numbers. If $\sM$ is countable, one can use this to construct an isomorphism $(\mathbf R^\sM,\mathbf Z^\sM,+,{<})\simeq(\mathbf R_{\mathbf L}^\sM,\mathbf Z_{\mathbf L}^\sM,+,{<})$, whose composition with $2^x$ yields the desired exponential on $\mathbf R^\sM$.

## References

[1] Emil Jeřábek: *Elementary analytic functions in $\mathit{VTC}^0$*, arXiv:2206.12164 [cs.CC], 55 pp., 2022.

[2] Emil Jeřábek: *Models of $\mathit{VTC^0}$ as exponential integer parts*, arXiv:2209.01197 [math.LO], 21 pp., 2022.

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