Why can't logarithms be expressed in terms of exponentiation?

Question

The inverse operation of addition is subtraction, and vice versa. They undo each other. Funnily enough, subtraction can be expressed in terms of addition. $a-b=a+(-b)$. Additionaly, the inverse operation of multiplication is division. Much like addition and subtraction, multiplication and division undo each other. And, also like addition and subtraction, division can be expressed in terms of multiplication. $a/b=a*(b^{-1})$. Moving one step further, the inverse of exponentiation is radication (also known as nth root). They undo each other, and, in a surprising turn of events, you can express a root in terms of a power. $\sqrt[a]{b}=b^{1/a}$ . Look at that! All fitting so nicely together, in perfect balance and harmony.

That last sentence was a lie. Of course not. This is math, for goodness sake.

I'm sure you are all aware that, due to exponentiation not being commutative itself, it actually has TWO inverses. Aside from roots, its SECOND inverse is logarithms. However, to the best of my knowledge, there is no way to express logarithms in terms of exponentiations like all the others do so nicely.

My question to you all is:

$1$) Why is there not a way to represent logarithms in terms of exponentiation?

$2$) If I've gotten it all wrong and there is, in fact, a way, what is it?

Answer 1

There is a way but it’s not as direct as you wish as the natural logarithm can be expressed as exponentiation $$\lim_{h\to0}\frac{x^h-1}{h} = \ln(x).$$ You must have faced this form while doing derivatives of exponential functions using the definition of a derivative formula and, since this is true, you can express any logarithm of whatever base exponentially by using the change-of-base formula $$\log_a b = \frac{\ln b}{\ln a}.$$ And I guess there is no other way to express it because if there were such a simple way then there would be no point in creating the logarithm function.

Answer 2

Since we're dealing with non-commutative operations, we have to be a bit more careful in how we phrase things. For simplicity, I'll use the latter half of the alphabet for variables when discussing addition, and use "negation" (as opposed to "subtraction") to refer to the unary function $p\mapsto -p$.

More substantively, infix notation will be useful for establishing the parallel I want here, so let's define some new notation:

• $a\rightarrow b$ and $b\leftarrow a$ will both mean "$b^a$" (we can motivate this by thinking of "functions from a set of size $a$ to a set of size $b$" in either case).

• Similarly, $a \triangleleft b$ and $b\triangleright a$ will both mean "$\log_b(a)$."

(We could also add infix versions of root-taking, but we won't need those here. Meanwhile, note that while addition is commutative, the above suggests that our usual notation is missing "reverse-subtraction" - we'll see that play an important role below!)

Let's look at how these play out. In the case of addition, we had $$x-y=x+(-y),$$ with binary-subtraction serving as a right inverse to addition in the sense that $(x+y)-y=x$. In the case of exponentiation, we have the following: $$(a\rightarrow b)\triangleleft b=\log_b(b^a)=a.$$ So if $\rightarrow$ is to be our $+$-analogue, then $\triangleleft$ should be our $-$-analogue.

OK, so that means that we want to find some $b^\circ$ (our analogue of $-b$) such that $a\rightarrow b^\circ=a\triangleleft b$. That is, we want to find a $b^\circ$ such that $(b^\circ)^a=\log_b a$. And this is easily solved to give the answer ... $b^\circ=\sqrt[a]{\log_ba}$? Wait, what?

Well, think back to the addition case! Suppose we forgot pretty much everything we knew about addition/subtraction/negation, and we wanted to solve for $y'$ in the equation $$x+y'=x-y.$$ All we know about subtraction in this case is that it satisfies $(u+v)-v=u$ for all $u,v$ (look back to the $\rightarrow/\triangleleft$ situation above). This isn't much to work with, especially if we want to distinguish between left and right versions of addition (which we need to since exponentiation isn't commutative). Writing "$\sim$" for "subtract from the left" (so e.g. $2\sim 5=3$), we have $$(*)\quad y'=x\sim(x-y).$$ Note the double appearance of $x$ here, paralleling the double appearance of $a$ in our surprising formula above. The fact that the expression $(*)$ can be simplified uses nontrivial facts about addition/subtraction which don't hold for the arrow-and-triangle operations we're playing with here.