For an integer number $a$

$$x^a=\{(x)(x)(x)...(x)\} (a\,times)$$ $$x^{\frac{1}{b}}=n\rightarrow\;\{(n)(n)(n)...(n)\}(b\,times)=x$$

For rational number $m=\frac{a}{b}$


And can be though of as a combination of the situations before

What about


How would one calculate or picture this from more basic operations?

Colin Hicks
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  • The definition for integer values or even rational values for the exponent are fine for building intuition, however the full definition for how exponentiation works is more complicated and does not rely on those naive introductory definitions. You should look up the formal definition of the function $\textbf{exp}(x)$ and $\ln(x)$. – JMoravitz Jul 08 '18 at 20:24
  • It can be described either as the limit of $x^{\frac{a_n}{b_n}}$ for any sequence of rational numbers $\frac{a_n}{b_n}$ converging to $\mathrm e$, or as the exponential of $\frac1n\ln x$. – Bernard Jul 08 '18 at 20:25

2 Answers2


$x^e$ is the limit of the sequence

$$x^2, x^{27/10}, x^{271/100}, x^{2718/1000}, \cdots$$

By the way. This is a conceptual not a computational definition. No one would want to compute the thousandth root of $x^{2718}$ by hand. Especially since the sequence will converge to $x^e$ very slowly.

According to Wolfram alpha, to the first ten digits...

$$5^e \approx 79.43235917 $$

$$5^{2718/1000} \approx 79.39633798 $$

$$\text{absolute error $= |5^e - 5^{2718/1000}| \approx 0.036$}$$

$$\text{relative error $= 100 \dfrac{|5^e - 5^{2718/1000}|}{5^e} \approx 0.045\%$}$$

Steven Alexis Gregory
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  • Is it obvious that a precision of $4$ digits on the input results in a precision of about $4$ digits on the output? – Arnaud Mortier Jul 10 '18 at 16:26
  • @ArnaudMortier No, it's not obvious. It's a rule of thumb used when you're too lazy to do the arithmetic. – Steven Alexis Gregory Jul 13 '18 at 16:12
  • Being lazy is something that I can understand, but this rule of thumb is not something I would use, even when lazy: in general the precision on the output can be much better, and it can be much worse, than that on the input, depending both on the function and on the input. Unless there is something about the function that makes it obvious. – Arnaud Mortier Jul 13 '18 at 16:26
  • Un this case, it has been pretty accurate. There are times when you need a BOTE estimate and being in the ballpark is good enough for now. The above method was only for conceptualization, it is a horrendous waste of time for practical purposes. – Steven Alexis Gregory Jul 13 '18 at 20:02

For any $a\in\Bbb R$ and any positive $x$, one has by definition $$x^a=e^{a\ln x}$$ While this could seem to be a loopy definition, it is actually not since $e^x$ is primarily defined not via exponentiation, but via one of the two equivalent definitions:

  1. $$e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$$

  2. $f:x\mapsto e^x$ is the only function $\Bbb R\to\Bbb R$ such that $f(0)=1$ and for all $x\in\Bbb R$, $f'(x)=f(x).$

Arnaud Mortier
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