# 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!

• $2^6=64=4^3=8^2.$ But $6^2, 3^4, 2^8$ are all different numbers. Commented Sep 27, 2022 at 3:32
• Are you expecting some function $f(c)$ on the other side? Consider $2^4=16$ and $4^2=16$. This implies that $f(16)$ would have to be $16$. But also $16^1=16$, and yet $1^{16}\neq16$. Commented Sep 27, 2022 at 3:32
• Means there is no general formula? Commented Sep 27, 2022 at 3:40
• The point is that knowing $a^b$ is not enough information to determine $b^a$. If you know $a^b=64$, then $b^a$ could be $81$ ($3^4$), $36$ ($6^2$), $256$ ($2^8$), or $1$ ($1^64$).
– Karl
Commented Sep 27, 2022 at 3:42
• If $a^b = c$, then $b = \log_a c$, so $b^a = (\log_a c)^a$. If that helps. Commented Sep 27, 2022 at 3:51

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

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 $$0 holds. Then, define the number $$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

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}

• No, I cannot determine. Thanks, I understand it now. Commented Sep 27, 2022 at 4:01
• A limitation exists also for the case $\frac{a}{b} = c$ when $a=0$ then $\frac{b}{a}=?$.
– user
Commented Sep 27, 2022 at 13:19
• @user The answer is expanded. Commented Sep 27, 2022 at 13:33
• @lonestudent But sir, I have assumed that $a$ and $b$ are constants and not variables. For example, $2^3=8$ but $3^2$ is unique because $2$ and $3$ are constants. Commented Sep 27, 2022 at 17:25
• @An_Elephant Added expanded and more detailed explanation. Commented Oct 3, 2022 at 17:22

$$c = a^b = e^{b\log(a)} \implies \log(c) = b\log(a) \implies$$

$$b = \frac{\log(c)}{\log(a)} \implies b^a = \left[\frac{\log(c)}{\log(a)}\right]^a.$$

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

$$\log_a \left[c^a\right] = a \times \left[\log_a(c)\right].$$

• Thanks . I cannot find reason but please tell that does this contradict the discussion above that $b^a$ is not unique. Commented Sep 27, 2022 at 9:53
• @An_Elephant No contradiction. All you really know about the values $a,b,c$ is that $a^b = c.$ You can not use this information, by itself, to determine a unique value for $b^a$. Commented Sep 27, 2022 at 11:12
• Means the unique value can only be determined only if $a$ and $b$ are constants? But I haven't assumed in the question that they are variable. So please explain. I am confused again. Commented Sep 27, 2022 at 17:22
• Please see the edited question. Commented Sep 27, 2022 at 17:51

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 $$a \neq 0$$.