I have two questions about logarithms that have been bugging me for years:

The first is regarding natural log. The domain of Ln(x) is (0, ∞). This makes sense since you can’t raise e to any power to get a negative number. However, I have seen the fact that Ln(x) and e^x used as inverse operations (in other words x =e^(ln(x)) ) in quite a few proofs. For example, here is one proving the power rule for derivatives for all real numbers using implicit derivatives from Paul’s Online Math notes:

Since Ln(x) only has positive numbers in its domain, shouldn’t this mean this proof is only valid when x is positive?

Another thing I have wondered about is what goes beyond roots and exponentiation. This is what I mean; the whole point of logs is that they turn addition & subtraction into multiplication & division, and multiplication & division into exponentiation and root extraction. For example (I’m using log to the base 10:

Log(64) = 1.8061799739838869
Antilog (3.612359947967774) = 4096 which is indeed the square of 64

We can go the other way too:
Log(64) = 1.8061799739838869
1.8061799739838869/2 = 0.903089986991943
Antilog(0.903089986991943) = 8 which is the square root of 64

My question is what happens when exponentiation and root extraction is extended as well. For example:
Log(64) = 1.8061799739838869
(1.8061799739838869)^2 = 3.262286098420434
Antilog(3.262286098420434) = 1829.3049030927973

Log(64) = 1.8061799739838869
(1.8061799739838869)^½ = 1.343941953353599
Antilog(1.343941953353599) = 22.077096366949473

I don’t know what I expected but I though the “super square” & “super root” of 64 would be something more meaningful. Am I missing something? (probably)

  • $\begingroup$ These should be broken into two questions. They are quite distinct from each other. The first is a good one, the second could use some more precision about what you mean by extending exponentiation and root extraction. I believe you are talking about tetration here without mentioning it. $\endgroup$ – Ross Millikan Aug 20 '18 at 3:33

You are correct that logs are only defined in the reals when the argument is positive. When we define $x^y$ with $x,y$ irrational it is usually as $e^{y \ln x}$ or something equivalent. This is also only defined for $x \gt 0$. In the reals we shouldn't talk of $\frac d{dx}x^y$ for $x \le 0$ because the function $x^y$ doesn't make sense for us to try to take its derivative. We can do $x^y$ for $x \lt 0$ if $y$ is an integer, and we can take its derivative with respect to $x$, but we shouldn't go through the log to get there. When we get to the complex field we can take logs of negative numbers, but they are multivalued.

  • $\begingroup$ Thank you for responding however I'm still stuck on it. Let me use an example to describe why. Suppose I cite the proof of the power rule using implicit differentiation cited in my question to differentiate x^2. Does that really mean that it only holds for positive numbers? Meaning can I really not claim the derivative of x^2 is 2x for negative numbers according to the implicit proof? I find it hard to believe that the implicit proof doesn't prove that the derivative of X^2 is -6 at x=-3. This probably means there is something very basic I'm not getting. Thank you. $\endgroup$ – Martino Ciaramidaro Aug 21 '18 at 22:01

Great question! The answer involves complex analysis.

First of all, log is defined to be the inverse of exponentiation. That is, $l=ln(x)$ is defined to be a value such that $e^l=x$. You stated that $e^x$ is never negative: this is true only when $x$ is restricted to be a real number. You may be familiar with the following famous identity:

$$e^{\pi i} = -1$$

(reference: https://en.wikipedia.org/wiki/Euler%27s_identity). Here we seem to have that $\pi i$ is the log of $-1$. Things get a little bit tricky here. If we square the identity we have

$$e^{2\pi i} = 1$$

Combining these we see that for any integer $k$:

$$1^k e^{\pi i} = e^{2k\pi i + \pi i} = -1$$

This is strange: it seems like $\ln(-1)$ is simultaneously $\pi i, 3\pi i, 5\pi i$ and so on. $\ln$ is called multivalued for this reason. There is a way of dealing with this so that $\ln$ is essentially uniquely defined, I would encourage you to read https://en.wikipedia.org/wiki/Branch_point to learn more.

As for your second question, it seems you are asking if anything can be said about $\exp(\log(x)^a)$. I do not know of anything particularly interesting. One thing you can note is that $e^{(\log(x)^a)}=x^{\log(x)^{a-1}}$.


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