How we calculate the value of $2^\pi$ without using calculator How we calculate $2^\pi$? since $\pi$ is irrational how shall I calculate this? and can we write $$(2^\pi)(2^\pi)=(2^\pi)^2$$ and if yes what will be the condition, since $\pi$ is irrational no.
 A: $$A=2^\pi=2^3\, 2^{\pi-3}=8 \, 2^{\pi-3}=8\, \exp\big[(\pi-3)\log(2) \big]$$ Now, for small $x$, a good approximation is
$$e^x=\frac{2+x}{2-x}$$ and $(\pi-3)\log(2)\sim 0.098$
$$A\sim 8 \,\frac{2+0.098}{2-0.098}=\frac{8392}{951}\sim 8.8244$$ while the exact result is $8.8250$.
A: If $n$ is a positive integer:

*

*$2^n$ means the product of $n$ copies of $2$,

*$2^{-n}$ means $(2^n)^{-1} = (2^{-1})^n$ (why this are the same number?),

*$2^{\frac 1n}$ means $\sqrt[n]{2}$, and

*$2^{-\frac 1n}$ means $(2^{\frac 1n})^{-1} = (2^{-1})^{\frac 1n}$ (again, why?).

Also, $2^0$ is $1$. Now, if $p$ and $q$ are two integers with $q \neq 0$, $2^{\frac pq}$ means $(2^{\frac 1q})^p = (2^p)^{\frac 1q}$ (?). With all of this, we know the value of $2^r$ for any rational number $r$, right? Finally, if $x$ is a real number, choose a sequence of rational numbers $r_1,r_2,\dots$ such that $\lim_n r_n = x$, and then $2^x$ means $\lim_{n} 2^{r_n}$.
A: *

*To answer your motivating question ("How can we visualise $2^\pi$? How we can understand it?") in the 4th comment under the main question:

$\\ a^x$ is defined as $e^{x\ln a}$ for all positive $a$ and real $x$.
$e^y$ is defined as $\lim_{n\to\infty}\left(1+\frac y{n}\right)^n$ for all real $y$.
Therefore, by definition, $$2^\pi=\lim_{n\to\infty}\left(1+\frac {\pi\ln 2}{n}\right)^n.$$
Equivalently (from the power series expansion of $e^{\pi\ln 2}$), $$2^\pi=\sum_{n=0}^{\infty}\frac{(\pi\ln 2)^n}{n!}.$$


*Plugging in the power series expansions of $(4\arctan1=)\pi$ and $\ln 2$ gives
\begin{equation}
\begin{split}
2^\pi=\sum_{n=0}^{\infty}\frac 1{n!}&\left[4\left(\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}\right)\left(\sum_{n=1}^{\infty}(-1)^{n+1}\frac1{n}\right)\right]^n \\
=1+\frac11&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right] \\
+\frac 1{2!}&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right]^2 \\
+\frac 1{3!}&\left[4\left(\frac 11-\frac 13+\frac 15-\ldots\right)\left(\frac 11-\frac12+\frac13-\dots\right)\right]^3 \\
+\ldots.
\end{split}
\end{equation}
This theoretically lets us estimate $2^\pi$; however it converges too slowly to be of practical use.

A: Here will make use of calculus approximation using differentials
$$ y= 2^x \\
\ln\ y = x\ln\ 2 \\
\frac{1\ dy}{y\ dx}\ = \ln\ 2 \\
dy = y\ln\ 2\ dx \\
dy = 2^x \ln\ 2\ dx \\
$$
now use a differential approach
$$
\Delta y = 2^x\ln 2\ \Delta x
$$
Now, using  $x=3, \ \Delta x = 0.14159265359$ we see that $\Delta y=78159144782$, that added to the original value in $2^3=8$ is $8.781581$ which is close to the value, although not as close as other methods shown here.
