An interesting exponent.

Let's just consider the $n = 2$ and $b > 0$ case: suppose that the function $f(x) = b^x$ is uniformly continuous when $b > 0$, then we can represent $\sqrt{2} = \lim x_k$ as the limit of some rational cauchy sequence. By uniform continuity, we can also claim that b^\sqrt{2} = f\left(\lim_k x_k\right)= \lim_k f(x_k) this means that we can construct a convergent sequence to b^\sqrt{2} by taking rational exponents, so for example, if we have nice technique/method of computing rational exponents, then we can compute 3^\sqrt{2} by computing $3^1, 3^{1.4}, 3^{1.41}, 3^{1.414}, 3^{1.4142},\dots$

From a computational point of view, the selection of the sequence matters because sequences that converges quicker will yield a better approximation. In the above, we can bound the error of $|x_k - \sqrt{2}| \le 10^{-k}$, so the relative error of the $k^{th}$ approximation is bounded above by $1 - b^{10^{-k}}$. Doing a bit of quick and dirty work (using a first order approximation of $b^x = x\log b + O(x^2)$, and let $\delta_k$ denote the relative error using the $k^{th}$ approximation, then for errors of the form $1 - b^{f(k)}$ $\delta_k \approx |f(k)\log b|$

For the sequence that adds in an extra decimal term each time, we can arbitrarily guarantee $m$ digits of accuracy in decimal point by allowing $10^{-m} \approx 10^{-k} \log(b) \implies k \approx m + \log \log b$

More interestingly, the following rational sequence approximating $\sqrt{2} = \lim_k x_k$ where the sequence is described by the recurrence $x_{k+1} = \frac{x_k^2 + 2}{2x_k} ~~~~ x_1 = \frac{3}{2}$ converges quite rapidly. In fact, within the region $x_k \in (1,2)$, it can be shown that $|\sqrt{2} - x_k| \le \frac{1}{2} \left(\frac{\sqrt{2} - 1}{2}\right)^{2^k}$ so here, to get m digits in decimal, you would only need to compute the $k = \log_2(m) - \log_{2}\log_{10}\left(\frac{2}{\sqrt{2}-1}\right)$, which is super quick!

We can write $b^{\sqrt{2}} = e^{\sqrt{2} \log b }$, from which it follows that if $b > 0$, the resulting function is well defined on the reals. When $b = 0$, then it is reasonable to interpret $0^{\sqrt{2}}$ as the limiting value of $f(x) = 0^x$ as $x \to \sqrt{2}$, which is of course $0$.

The interesting that happens is when $b < 0$. Can we formulate an axiomatically consistent interpretation of the function $g(b) = b^{\sqrt{2}}$ for $b < 0$? Can we do so for non-real $b$? The above suggests employing the complex logarithm: $\log z = \log |z| + i \arg z + 2 \pi i k, \quad k \in \mathbb Z,$ where $\arg z$ is the complex argument of $z$; i.e., the angle that $z$ makes with the positive real axis in the complex plane.

With this in mind, we observe that for general complex $b$, the expression $b^{\sqrt{2}}$ is a set of complex numbers. We can choose a particular branch for this multivalued function to make it well-defined, but the result will contain a branch cut, across which the value of the function will be discontinuous.