Pragmatically
The short answer is that, almost without exception, in a pragmatic sense, we often need inverse operations. Consider, for example, something you would solve in your introduction to algebra.
$$2x = 6$$
This question is entirely about inversion. In words it asks: "Knowing that something when multiplied by two is equal to six, determine the original value."
Answering this question requires that we talk about inverse operations, how can we define something that undoes multiplication and what are its properties?
Maps and Multiplication
But there is a deeper reason that operations occur in pairs (small caveat, we will come back to this). An operator can be thought of as a mapping from one set of numbers, or a set of tuples of numbers, to another set. For example, on the set of inputs $\{1, 2, 3\}$ we can exhaustively define the map produced by multiplication. Let $m(x_1,x_2)$ be a map from $\{1,2,3\}^2\to\mathbb{N}$ defined as normal multiplication $x_1x_2$.
\begin{align}
m(1,1) \mapsto 1 & \quad m(1,2) \mapsto 2 & m(1,3) \mapsto 3 \\
m(2,1) \mapsto 2 & \quad m(2,2) \mapsto 4 & m(2,3) \mapsto 6 \\
m(3,1) \mapsto 3 & \quad m(3,2) \mapsto 6 & m(3,3) \mapsto 9
\end{align}
Can we define a mapping that inverses this operation? Of course! Just point the arrows the other way.
\begin{align}
1 \mapsto m(1,1) & \quad 2 \mapsto m(1,2) & 3 \mapsto m(1,3) \\
2 \mapsto m(2,1) & \quad 4 \mapsto m(2,2) & 6 \mapsto m(2,3) \\
3 \mapsto m(3,1) & \quad 6 \mapsto m(3,2) & 9 \mapsto m(3,3)
\end{align}
Now, you may notice that our mapping has a problem, $2$, $3$, and $6$, all map to more than one object. This mapping is not proper because the original mapping (multiplication) was not injective.
This should come as no surprise however, since we know that unless a number $n$ is prime we can't find a unique solution to
$$n=x_1x_2$$
with $x_1$ and $x_2$ being positive integers. (And even then, we cannot say which of the two is $1$ and which is $n$.)
But when we do division, and solve the equation mentioned above, $2x=6$, we are not working with both $x_1$ and $x_2$ unknown. We are looking for which of the expressions above we can match to $6\mapsto m(2,x)$, You will notice there is only one such entry in the list above, and it has $x=3$ as the second parameter.
It turns out that for division, the 'question' $6\mapsto m(x,2)$ gives the same solution, and that this is the case for all $x_1$ and $x_2$, and so multiplication and division come in what appears to be a pair.
A tiny bit more formally, for a number $n$, the improper map $m^{-1}$ from $\mathbb{N}\mapsto\{1,2,3\}^2$ can be made proper in two ways. We can fix the first value in the tuple, or the second. $\mathbb{N}\mapsto(x_\text{fixed},x_2)$ or $\mathbb{N}\mapsto(x_1,x_\text{fixed})$. These two ways of making the inverse mapping proper happen to be identical.
Asymmetric Inverses: Exponentiation
This is not always the case. You mention logarithms and exponentiation. Let's do a similar exercise, consider $e(x_1,x_2) = {x_1}^{x_2}$ in the same way as before, and let's investigate its behaviour on some choice examples:
\begin{align}
e(2,2) \mapsto 4 && e(2,4) \mapsto 16 && e(2,8) \mapsto 256 \\
e(4,2) \mapsto 16 && e(4,4) \mapsto 256 && e(4,8) \mapsto 65536 \\
e(8,2) \mapsto 64&& e(8,4) \mapsto 4096 && e(8,8) \mapsto \text{nope}
\end{align}
This is different from $m$ in a big way. Sometimes it plays nice and we have ${x_1}^{x_2}={x_2}^{x_1}$, but sometimes (mostly) that's not the case. The inverse is a perfectly acceptable improper map, but the inverse in the case that one of the arguments of $e$ is known is different depending on the known argument.
The logarithm is only one of the proper inverses that exponentiation has: if we fix the base and let $x_1=b$, so we are inverting $e(b,x_2)$, then we obtain the familiar logarithm $$e^{-1}_{x_1 = b}(n) = \log_b(n)$$
If we fix the exponent, however, we get another reasonable inverse to exponentiation: rooting.
$$e^{-1}_{x_2 = p}(n) = \sqrt[p]{n}$$
This is just as valid an inverse to exponentiation as logarithms. Where logarithms ask "to what power must a given base be raised to obtain a given result", $x_2$-th/$p$-th rooting asks "what base when raised to a given power yields a given result?" That is, in $b^p=n$ the logarithm base-$n$ makes $p$ the subject and $p$-th rooting makes $b$ the subject.
$$\log_b(n) = p \qquad \sqrt[p]{n} = b$$
Summary
Unary operations, like successorship, negation, and trigonometric functions, will always have a unique inverse. This 'symmetry' as you call it is a result of the fact that we can always invert the mapping an operation represents and then ask about its behaviour.
Binary operations, like multiplication, exponentiation will always have two inverses. It may appear as though there is only one, but this is a result of the fact that some of the most basic operations produce a pair of inverses which is identical. In the case of exponentiation, you simply never learned a name for the second inverse and so incorrectly assumed that there was only one.
$n$-ary operations will, thus, have $n$ inverses, with $m \leq n$ unique inverses.
