# How to calculate a decimal power of a number

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.

• 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. Dec 6, 2013 at 9:58
• @arbautjc I would only need the first 5 decimals, I guess it would be possible in finite steps then. Dec 6, 2013 at 10:10
• 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 ;-) Dec 6, 2013 at 10:14
• @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. Dec 6, 2013 at 10:16
• @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). Sep 24, 2014 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).

• 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. Feb 10, 2011 at 16:24
• @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 $(2.14)^{107}$. Feb 10, 2011 at 16:34
• 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. Dec 6, 2013 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.

• 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 Feb 10, 2011 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

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

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

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

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

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

• I like your answer. But requiring a calculator for the "intermediate step" kinda defeats the purpose. Dec 19, 2013 at 8:58
• +1. I reached this conclusion by trial and error and didn't understand why it worked. Beautiful explanation. Apr 17, 2015 at 15:27
• 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. Oct 14, 2019 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 $\ln 2$.

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.