The underlying issue here is that (assuming you want to stay within the real numbers) when $c<0$, the function $c^x$ is undefined for most values of $x$. Specifically, it's undefined unless $x$ is a rational number whose denominator is odd. There is no continuous/differentiable function underlying the places where it is defined.
Therefore, there is no possible guess-and-check algorithm that gradually becomes more accurate. First, guess-and-check algorithms require an underlying continuous function. Second, the value you're seeking might simply not be defined.
So the need to determine whether the exponent is a fraction with odd denominator, which in other contexts might be considered inelegant, here is simply a necessary step in the problem you're trying to solve. (And really, people shouldn't be inputting $c^x$ when $c<0$ and $x$ is a decimal ... they're just asking for trouble, for all the reasons mentioned above.)
Suppose you want to compute ${3.21}^{1/5}$. If you have a logarithm table (say base 10), you only need the logarithms of numbers between 0.1 and 1 stored (alternatively between 1 and 10), as many as is relevant for your precision.
Then because
$$\log (3.21^{1/5}) = \frac{1}{5}\left(\log(10) + \log(0.321)\right)= \frac{1}{5}\left(1+\log(0.321)\right)$$
Now you look up $\log(0.321)$ in your table, which will look something like this
$$\begin{array}{c|c}
\text{Input} & \text{Output} \\
\hline
\ldots & \ldots \\
\color{red}{0.321} & -0.493 \\
\ldots & \ldots
\end{array}$$
do the above computation
$$\frac{1}{5}(1+\log(0.321)) = \frac{1}{5}(1-0.493) = 0.101$$
and look up the result in the "answer column" of your table to revert the $\log$ operation. Since the answer is positive, and we worked with a table containing logarithms of numbers between 0 and 1, we'll need to look up the opposite first
$$\begin{array}{c|c}
\text{Input} & \text{Output} \\
\hline
\ldots & \ldots \\
0.792 & \color{red}{-0.101} \\
\ldots & \ldots
\end{array}$$
and now take the inverse of that number to obtain the result: $1.262$.
Best Answer
You follow the rule $$b^{-n} = \frac{1}{b^n}$$ Proving the rule is pretty simple ($b\neq 0$): $$b^n \cdot b^{-n} = b^{n-n} = b^0 = 1 \\ \implies b^n \cdot b^{-n} = 1 \implies b^{-n} = \frac{1}{b^n}$$
Edit: for decimals, it really depends on what the decimal is. For example $4^{0.5}$ is simple because $0.5=1/2$ thus $4^{1/2}=\sqrt{4}=2$. But if you had something tougher like $4^{1.5}$ then the method is similar: $4^{1+0.5} = 4^1 \cdot 4^{1/2} = 4\cdot 2 =8$ But when you make it tougher, like $4^{1.234}$ then it gets tedious to split it up the way we did and write them as fractions and so on. Sometimes you end up with things like $4^{1/3}$ which is the cube root of 4. You can only approximate such things. So unless it's a simple one, one would usually use a calculator.