On the super logarithm and interpolation to real inputs

So, I think most of you are familiar with Tetration, the hyper operations, and logarithms. But one flaw of these is that you can't extend them to real inputs. What if I want to find the result of a power tower of π 3s? What if I want to create a function so slow, Inputting a 5 entry BEAF array will only output 2? well, I think I have a somewhat boring answer.

Okay, but what ARE hyper operations?

I think you guys know how they work, but how exactly are they defined? The hyper operators are defined using the rules:

• $a\uparrow^{1}b=a^b$
• $a\uparrow^{n}1=a$
• $a\uparrow^{n+1}b = a\uparrow^n (a\uparrow^{n+1}(b-1))$

in this case, I've used the most common notation, the Knuth up-arrows.

Now, before we can tackle expressions like $1.15 \uparrow^{6.8} 5.511$, let's just look at each individual part: the base (first number), the prime (last number), and the index (superscript on the arrow). Lets do them in order from easiest to hardest.

Exhibit 1: Base not an integer

This is trivial. The only problem is by reducing the number (say, 1.6^^^3) we will bump into the problem of having to deal with fractional primes.

On the stupidness of exponents

You might be wondering, "how do regular logarithms relate to the hyper-operations? I mean, even numbers like 5^^^5 are pretty much untouchable using things like logarithms." And you'd be right. However, just like you can reverse-engineer expressions like $5^{\pi}$using logarithms, you can do the same thing for hyper operations. One problem is that there aren't any nice properties such as $a^ba^c = a^{b+c}$, or is there? This is where we have to take a moment to rant about why exponents suck.

The fact is that despite making math much simpler, exponents aren't commutative, and so there is 2 inverse operations, roots and logarithms. logarithms are far more powerful, because you essentially just drop the base from the number, whereas for roots, you just divide the exponent, leaving the base intact. To show you what I mean, here's an example: $\log_5(15625) = 6 < \sqrt[5]{15625} \approx 6.8986\cdots$in this case, the first example is the map of 5^6 -> 6, where the second is 5^6 -> 5^(5/6). However, this also satisfies the a^b+a^c = a^(b+c) property from earlier, and also satisfies $(a^b)^c = a^{b\times c}$, another nice property. This doesn't excuse the fact that I failed my math test due to mixing these up, but let's get back on track and start defining logarithms for higher operations.