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Inverse of exponentiation

  1. Feb 27, 2009 #1
    What is the exponential inverse? And first let me explain what I mean when I am talking about inverse.

    I am not talking about anything where f(f-1(x)) = x

    What I mean is in one operation, when that operation is applied onto the inverse of that operation it equals the identity element of that operation.
    For example, 5 + -5 = 0 because 0 is the identity element in addition (x+0=x), and negative is the inverse.
    Also for example, 5 * (1/5) = 1 because 1 is the identity element in multiplication (x*1=x), and reciprocal or 1/x is the inverse.
    In matrices, [matrix] * [inverse of that matrix] = [identity matrix]

    So, what is this for exponentiation?

    Let me define what I want to use for identity element: anything, that when that operation is applied to it, it equals the original number. So in exponentiation, the identity element is 1 because x^1 = x;

    So now, we want x^what = identity element; And since the identity element of x is 1 then this can become: x^what = 1?

    I have considered 0 as the answer, but that doesn't make any sense. Even though x^0 = 1, and thus the identity element, how is 0 the exponential inverse of x? All of the other inverses for addition and multiplication (-x, 1/x) all include x. So why, here, is 0 the inverse; what makes exponentiation and 0 so special?

    In addition, root and logarithm do not work. Root doesn't work because x^(x^(1/anything)) does not equal 1 (unless x = 1). and neither does x^logx(anything)

    So x^what = 1, where "what" can't be zero, a root, or a logarithm.
  2. jcsd
  3. Feb 27, 2009 #2
    Oh and let me also add that taking a higher operation, and applying the negative, will result in the lower operation's inverse.

    x * -1 = the additive inverse = -x
    x ^ -1 = the multiplicative inverse = 1/x

    So to find the exponential inverse, then that would be the negative tetration (the operation higher than exponentiation).
  4. Feb 27, 2009 #3
    In the system you have defined y = 0 is the only solution to x^y = 1 so that is all you have.
  5. Feb 28, 2009 #4


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    How are you thinking of exponentiation? xn as a function of x for fixed n? Or ax for fixed a? If the former then the inverse function is the nth root. If the latter then it is the logarithm base a.
  6. Feb 28, 2009 #5
    Let me explain why I think log and root don't work for this, but first let me reiterate some things:

    In addition and multiplication, the inverses have this property: When they are added (or multiplied) to the original, non-inverted number, they equal the identity element (0 and 1 respectively).

    So if we try to apply this using exponentiation, then how can we do that?

    [x] [operation] [inverse x] = [identity element]
    x + (-x) = 0
    x * (1/x) = 1
    x ^ [exponential inverse x] = 1

    what is exponential inverse of x that will fit here? (except 0)

    While it is true, that to get the original x, root and logarithm can apply. sqrt(x^2)=x and logarithm works too. Allow me to explain my reasoning below.

    The reason the inverse function works in addition and multiplication is because of the original property I defined above, where doing that operation on the inverse will equal the identity element. So for x+5, to get x you would make it x+(5+(-5))=x. In multiplication, x*5, you would do x*(5*(1/5))=x. Following this, then for x^2, to get x you would need to do sqrt(x^2); seems pretty obvious. But notice how in the examples for multiplication and addition, the inverse we are doing never touches x. The 5 and 1/5 cancel out, and the 5 and -5 cancel out. x is left alone. But here, the square root DOES affect x, and so this doesn't follow the pattern. What I mean is that the square root is ((x^2)^1/2). Here, you can just multiply 2 and 1/2 and they become 1, thus making it x^1=x; Great. It all works out fine. So what is the problem? Well, we are not exponenting 2^1/2, but we are exponenting the entirety of x^2, which we will then root. But we want x^(2^(whatever)). Essentially, in multiplication and addition, x is not affected when the inverse is done; In exponentiation, to get the original x, you need to affect x. What I want, is a number [defined as "whatever"] which only affects the 2. The only solution to this is 0, because x^2^0 = x^1 = x; Ok, that's great. We found it, it is 0. Now where's the problem? We want something other than 0, since 0 is not related to x; it is the same for all x; Whereas -x and 1/x include x, 0 does not include x;

    Do you kind of understand my point a bit better?
    Last edited: Feb 28, 2009
  7. Feb 28, 2009 #6
    You should realize that in [tex] \mathbb{R} [/tex] the exponential function [tex]f(t) = a^{t} [/tex] is injective, so there is only one t for which [tex] a^{t} = 1 [/tex], which is 0. So in the 2-variable function [tex] g(x,y) = x^y [/tex], if and only if y = 0 does [tex] g(x,y) = 1 [/tex], whatever the choice of x. Even in [tex] \mathbb{C} [/tex] it doesn't make a difference because it will only work for [tex] y \in \{2{\pi}k : k \in \mathbb{Z} \} [/tex] which does not 'depend' on x either.
  8. Feb 28, 2009 #7
    Not true, it would be (x^2)^0 = 1. You have to apply it to the entire function. This doesn't matter with addition/multiplication because they're communicative but it does here.

    This is because addition commutes. What you're actually doing is
    x+5 = k
    (x+5) + (-5) = k + (-5)
    x+(5+(-5)) = k + (-5)
    x + 0 = k + (-5)
    x = k + (-5)

    No number fits your description. For real X and n, the only way to have X^n = 1 is to have n=0 (and [itex]X \not= 0[/itex]).
    Last edited: Feb 28, 2009
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