0
$\begingroup$

Here is my thought process:

  1. We can define subtraction as adding the inverse of a number w.r.t. the addition operation, e.g. $f(x) = x - 2$ is equivalent to taking $x$ and adding to it a number $a$ such that $a + 2 = 0$, and we call $a$ “minus 2”.
  2. We can define division as multiplying with the inverse of a number w.r.t. the multiplication operation, e.g. $f(x) = \frac{x}{2}$ is equivalent to taking $x$ and multiplying it by a number $a$ such that $a * 2 = 1$, and we call $a$ “1 over 2".
  3. The same applies to roots as the inverses of raising to a power, although here things already start to get a bit weird, since we can think of $f(x) = \sqrt{x}$ as raising $x$ to a power $a$ such that $a * 2 = 1$. This is peculiar because in order to find the inverse of the raising to a power operation, we make use of the rules of multiplication (which arise from the rule that ${x^a}^b = x^{a * b}$, but still seems peculiar).
  4. With exponentiation this breaks apart completely. For example, $f(x) = \log_{2}{x}$ cannot be written as “exponentiating” $x$ in any way. It can be explained as taking $2$ and raising it to an exponent containing $x$ such that the end result is $x$, and we call that exponent “the logarithm base 2 of x”. However, this way of explaining things is conceptually different from cases 1-3.

I would assume those observations could be explained through abstract algebra, which would explain why the raising to a power operation relies on the properties of multiplication, and why exponentiation is in a category of its own, but I don’t know enough about this field to make those connections.

$\endgroup$
3
  • 1
    $\begingroup$ For 4th case, I would say you can vaguely define as such a number $a$ such that $2^a$ is $x$. $\endgroup$ Commented Sep 11 at 13:54
  • 2
    $\begingroup$ Hi, welcome to Math SE. Since exponentiation doesn't commute ($2^3\ne3^2$), there are two different ways to invert it. This video may help you understand what's going on. $\endgroup$ Commented Sep 11 at 13:59
  • $\begingroup$ If you want, you can use a "symmetric exponentiation", $e^{\log(a)\log(b)}=a^{\log(b)}=b^{\log(a)}$. $\endgroup$ Commented Sep 11 at 14:19

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.