How to calculate a decimal power of a number

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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

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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

Answer 4

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

Answer 5

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 $\ln 2$.

Answer 6

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.