Can we express the value of $b^a$ in terms of $c$ , where $c=a^b$? We know that ;
If $a+b = c$, then $b+a = c$
If $a-b = c$ , then $b-a=-c$
If $ab = c$ then $ba = c$
If $\dfrac{a}{b} = c$ then $\dfrac{b}{a} = \dfrac{1}{c}$
Now, I am curious to know that If $a^{b} = c$ , then what is $b^a$ in terms of $c$ ?
I tried using logarithms but failed to get the desired answer in terms of $c$.

EDIT: It has been made clear from the answers that $b^a$ is not unique because $a^b$ can also be expressed as $x^y$ where $x,y$ can take infinite values. However, what happens if $a$ and $b$ are not changed? I want to emphasize more on the numerical value of $b^a$ rather than preserving numerical value of $a^b$ and then changing it to for some other forms of $x,y$ such that $x^y=c$.
For example, $2^6 = 64$ and $6^2=36$ which is unique. It is clear that $64 = 4^3$ ;but we do not wish to change $a$ and $b$ by other possible values which also happen to be some solutions of the $c$.
Or in more mathematical terms, I am intrested in ;
If for some $a,b$ we have :
$a^b = c$ then find $y^x$ given that :
1)$x^y = c$ as well as
2)$x=a$ , $y=b$
If it is impossible to find, then please also provide a proof . Thanks!
 A: Consider $a=1$. Then you have: $a^b=c=1$ holds $\forall b\in\mathbb R$.
Can we determine the value of $b^a=b^1=b\,?$

If you're looking for a "more mathematical" answer with rigorous constraints, the following method might work for us.
Let $a$ and $b$ are real numbers, such that
$$0<a\neq 1\wedge 0<b\neq 1.$$
Then $\forall m>0$, you have
$$a^b=\left(a^m\right)^\left(\frac bm \right)=c$$
This implies that,
$$b^a:=\left(\frac bm \right)^{a^m}$$
Then, can the value of $\left(\frac bm \right)^{a^m}$ be unique $\forall m>0\,?$
Thus, you have shown that
$b^a$ takes infinitely  many number of different values.

Expanded and more detailed explanation:
If we cannot define your problem more precisely, we cannot offer a solution.
In this context $a$ and $b$ are fixed real numbers, such that $0<a\neq 1\wedge 0<b\neq 1$ holds. Then, define the number $c$ as $c:=a^b$. We need to find the algebraic function $f:\mathbb R^{+}\longrightarrow \mathbb R^{+}$, such that $f(c):=b^a$ holds. If we define the function $y:=f(c)$, then we consider $c$ as a variable.
Otherwise, your question will lose its mathematical meaning if we don't consider $c$ as a variable. For instance:
$$c=2^3, \; f(c):=3^2$$
or
$$c=5^8,\; f(c):=8^5$$
Observe that, this is obviously impossible to construct the function $f:\mathbb R^{+}\longrightarrow \mathbb R^{+}, \,f(c)$  such that $f(c):=b^a$, where $f(c)$ is an algebraic function, that doesn't involve any of the numbers $a$ or $b$.
Proof: Applying the same argument $\forall m>0$ we have,
$$
\begin{align}&\begin{cases}f(c):=f\left(a^b\right)=b^a\\f(c):=f\left(\left(a^m\right)^\left(\frac bm \right)\right)=\left(\frac bm \right)^{a^m}\end{cases}\\
\implies &b^a=\left(\frac bm \right)^{a^m},\forall m>0\\
&\text {A contradiction.}\end{align}
$$
In other words,
$f(c)\equiv b^a\equiv\left(\frac bm \right)^{a^m}$ is not unique.

Conclusion:

You can not construct the function $f:\mathbb R^{+}\longrightarrow \mathbb R^{+}, \,f(c)$  such that $f(c):=b^a$, where $f(c)$ is a standard mathematical function, that doesn't involve any of the numbers $a$ or $b$.

Remember that, you can easily construct the bivariate function $f(a,c)$ using logarithms.  But,  you can also construct the equivalent function $g(b,c)$ without using logarithms:
$$
\begin{align}&c=a^b\wedge 0<a,b\neq 1\\
\implies &a=c^{\frac 1b}\\
\implies &b^a=b^{c^{1/b}}
\end{align}
$$
Thus, your function can be defined as follows:
$$g:A \longrightarrow \mathbb R^{+},\;\;\;g(b,c):=b^{c^{1/b}}$$
where, $\begin{align}A:=&\left\{(b,c){\large{\mid}} b\in\mathbb R^{+}\setminus \left\{1\right\},\;c\in\mathbb R^{+}\right\}\end{align}$
A: $$c = a^b = e^{b\log(a)} \implies \log(c) = b\log(a) \implies $$
$$b = \frac{\log(c)}{\log(a)} \implies b^a = \left[\frac{\log(c)}{\log(a)}\right]^a. $$

Alternative expression is that
$$c = a^b \implies b = \log_a(c) \implies b^a = \left[\log_a(c)\right]^a. \tag1 $$
Note
The expression in (1) above is different from
$$\log_a \left[c^a\right] = a \times \left[\log_a(c)\right].$$
A: We have that for $a>0$ and $a\neq 1$, $b>0$ and $c>0$
$$a^b=c \iff b=\log_a c \iff b^a =\left(\log_a c\right)^a$$
otherwise it is not invertible in general, as for
$$c=\frac a b \iff \frac b a = \frac 1 c$$
wich requires $b\neq 0$ and $a \neq 0$.
