How to calculate a decimal power of a number

For positive bases $a$, you have the general rule $$a^b = \exp(b\ln(a)) = e^{b\ln a}.$$

This follows from the fact that exponentials and logarithms are inverses of each other, and that the logarithm has the property that $$\ln(x^r) = r\ln(x).$$

So you have, for example, \begin{align*} (2.14)^{2.14} &= e^{\ln\left((2.14)^{2.14}\right)} &\quad&\mbox{(because $e^{\ln x}=x$)}\\ &= e^{(2.14)\ln(2.14)} &&\mbox{(because $\ln(x^r) = r\ln x$)} \end{align*} Or more generally, $$a^b = e^{\ln(a^b)} = e^{b\ln a}.$$

In fact, this is formula can be taken as the definition of $a^b$ for $a\gt 0$ and arbitrary exponent $b$ (that is, not an integer, not a rational).

As to computing $e^{2.14\ln(2.14)}$, there are reasonably good methods for approximating numbers like $\ln(2.14)$, and numbers like $e^r$ (e.g., Taylor polynomials or other methods).

A decimal power can be seen as a fraction:

$x^{\frac{a}{b}} = \sqrt[b]{x^a}$

Of course you cannot write every number as a fraction, but you can at least approximate every number by a fraction.

we can find $2.14 ^{2.14}$ using basic arithmetic operations +,-,/,*.

Use binomial theorem for rational number $n and $-1

$(1+x)^n =1+nx+n(n-1)x^2/2!+....$

note that in left hand side the power n is a fractional number but in the right hand side the powers are integers. that is, in the right hand side, each term can be calculated using basic operations +,-,*,/.

outline of the problem

$2.14^{2.14}=(1.14+1)^{2.14}$

$=(1.14^{2.14}) * (1+1/1.14)^{2.14}$

$=(1+0.14)^{2.14} * (1+1/1.14)^{2.14}$

using binomial theorem two times (5 decimal places) and multiplying we get the answer