I wish to calculate a power like $$2.14 ^ {2.14}$$

When I ask my calculator to do it, I just get an answer, but I want to see the calculation.
So my question is, how to calculate this with a pen, paper and a bunch of brains.

  • 1
    $\begingroup$ By hand, you will need a table of logarithms or a slide rule. This operation can't be done in finitely many steps, like ordinary addition or multiplication, you have to resort to approximation. $\endgroup$ – Jean-Claude Arbaut Dec 6 '13 at 9:58
  • $\begingroup$ @arbautjc I would only need the first 5 decimals, I guess it would be possible in finite steps then. $\endgroup$ – Mixxiphoid Dec 6 '13 at 10:10
  • 1
    $\begingroup$ You are right, but it's still an iterative process that is not very easy, and very time consuming. Historically, it was done to build tables, which were then used for any computation of this kind. By the way, 5 decimals is the precision given by "standard" tables of logarithms ;-) $\endgroup$ – Jean-Claude Arbaut Dec 6 '13 at 10:14
  • $\begingroup$ @arbautjc thanks for the helpful links and comments. I will study those links further, I'm curious to the methods they applied to create the table. $\endgroup$ – Mixxiphoid Dec 6 '13 at 10:16
  • $\begingroup$ @Mixxiphoid You can compute the natural logarithm and the exponential from their power series (though there may be more efficient methods). The really magical thing is that you don't need log tables to use a slide rule, since they're essentially built in. Somebody with a slide rule who is still in practice (not so many people these days) should be able to demonstrate $2.14^{2.14}$ quite quickly (though maybe not to five decimal places). $\endgroup$ – Slade Sep 24 '14 at 21:03

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).

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks for the effort! I hoped for a bit more easy way to calculate is, but that seems to be impossible. I'll accept your answer. $\endgroup$ – Mixxiphoid Feb 10 '11 at 16:24
  • 9
    $\begingroup$ @Mixxiphoid: For rational exponents, you can also use the fact that $a^{1/b} = \sqrt[b]{a}$. Here, you have $2.14 = \frac{214}{100} = \frac{107}{50}$, so you could "just" take the 50th root of $(2.14)^{107}$. $\endgroup$ – Arturo Magidin Feb 10 '11 at 16:34
  • 1
    $\begingroup$ Slightly easier if you write $50=2\cdot 5^2$, but fifth root is not that easy either. For square root, on the other hand, there is an algorithm that resembles a bit the usual euclidian division. $\endgroup$ – Jean-Claude Arbaut Dec 6 '13 at 10:06

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.

| cite | improve this answer | |
  • 2
    $\begingroup$ Moreover, for $y \in \mathbb{R}\setminus\mathbb{Q}$, you can define $x^y := \sup \{x^r \;|\; r \in \mathbb{Q}, r \lt y\}$ which admittedly is only really useful when one is trying to develop the fundamentals of real analysis from scratch $\endgroup$ – kahen Feb 10 '11 at 17:03

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


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

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

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

| cite | improve this answer | |

I really like your question. So many students are content with learning (and so many instructors content with teaching) just the calculator key sequences that will give the correct answer. But to know math (and almost everything else in this world) you’ve got to get under the hood and “see” what’s actually going on.

Let’s start with your example, 2.14^2.14. When you look at the exponent, more than likely you intuitively get the feeling that one portion of the answer is due to the integer part, “2”, with the balance attributable to the decimal “0.14”. And you’re right.

So, let’s raise 2.14 to our integer power (which you can do by hand— though there’s nothing wrong with employing a calculator when you understand the manipulations you’re carrying out):
2.14 ^ 2 = (2.14 * 2.14) = 4.5796.

Actually, let’s back up a little and use our calculator to get the answer to our example; 2.14 ^ 2.14 = 5.09431.

Now that we have ‘the answer’ and the portion attributable to the integer component of our exponent, let’s determine the increase contributed by our decimal component; (5.09431/4.5796) = 1.112392. Ok, but other than the ratio, (5.09431/4.5796), just what is “1.112392”?

Fasten your seat belt— It is simply 2.14 ^ 0.14 power = 1.112392.
(Yes, use your calculator for this intermediate step)

So, 2.14 ^ 2.14 = (2.14 ^ 2 * 2.14 ^ 0.14) = (4.5796 * 1.112392) = 5.09431

Let’s try 5.27 ^ 4.34 = 1357.244436

 5.27 ^ 4 = 771.33397441…   5.27 ^ 0.34 = 1.759607

 (771.33397441 * 1.759607) = 1357.244436

Hope this is what you were looking for. Have fun! JE Magee

| cite | improve this answer | |
  • 4
    $\begingroup$ I like your answer. But requiring a calculator for the "intermediate step" kinda defeats the purpose. $\endgroup$ – StackedCrooked Dec 19 '13 at 8:58
  • $\begingroup$ +1. I reached this conclusion by trial and error and didn't understand why it worked. Beautiful explanation. $\endgroup$ – Lefty Apr 17 '15 at 15:27
  • $\begingroup$ You still relies on the calculator and don't know the principle behind the calculator. Basically what you did was just writing $a^{(b+c)}=a^b*a^c$, which is trivial. $\endgroup$ – Sam Wong Oct 14 '19 at 7:10

You use $\exp(2.14 \ln 2.14)$ or any base for logarithms you choose. But if you want pen and paper, you can help with the properties of exponents. $2.14^{2.14}=2.14^2\cdot2.14^{.14}=2.14^2\exp(.14(\ln 2 + \ln1.07))$ will converge more quickly, especially if you are willing to look up $\ln 2$.

| cite | improve this answer | |

Newton's approximation for $r = \sqrt{c}$ gives the iteration $r_{n+1} = r_n - \frac{{r_n}^2-c}{2r_n}$
$\sqrt{2.14} \approx 1.5 \rightarrow 1.46 \rightarrow 1.4628 \rightarrow 1.462874 \text{ (6sf)}$
Using that $10$ times gives $2.14 \rightarrow 1.462874 \rightarrow 1.209493 \rightarrow 1.099769 \rightarrow 1.048698 \rightarrow 1.024059$
$\rightarrow 1.011958 \rightarrow 1.005961 \rightarrow 1.002976 \rightarrow 1.001486 \rightarrow 1.000743 \text{ (6sf)}$
Thus $\ln 2.14 = 2^{10} \ln 2.14^{2^{-10}} \approx 2^{10} \ln 1.000743 \approx 2^{10} \times 0.000743 \approx 0.7608 \text{ (3sf)}$
$2.14^{2.14} = e^{ 2.14 \ln 2.14 } \approx e^{ 2.14 \times 0.7608 } \approx e^{1.628} \text{ (3sf)}$

The geometric series or binomial expansion gives the approximate
$2^{-10} = (1000+24)^{-1} \approx 1/1000 - 24/1000^2 + 576/1000^3$
Thus $e^{1.628} = (e^{1.628 \times 2^{-10}})^{2^{10}} \approx (e^{0.001590})^{2^{10}} \text{ (3sf)}$
$\approx (1+0.001590+0.001590^2/2)^{2^{10}} \approx 1.001591^{2^{10}} \text{ (6sf)}$
Squaring $10$ times gives $1.001591 \rightarrow 1.003185 \rightarrow 1.006380 \rightarrow 1.012801 \rightarrow 1.025766 \rightarrow 1.052196$
$\rightarrow 1.107116 \rightarrow 1.225706 \rightarrow 1.502355 \rightarrow 2.257071 \rightarrow 5.094369 \approx 5.09 \text{ (3sf)}$

which is $2.14^{2.14}$ to $3$ significant figures. I am lazy so I used a calculator for nine of the repetitions of square-root and squaring, but the above computation is clearly feasible by hand as only $O(n^3)$ operations are needed for $n$ bits of precision. It is amusing that so much work went in to produce only 3 decimal digits but I do not know any better way that can be easily extended to arbitrary precision.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.