There might be no "answer" but I figured it was worth asking the community.
NOTE I wanted to share what ChatGPT thought about this, ChatGPT responded with:
The logarithmic formula $\log_{b} x = \frac{\log_{k} x}{\log_{k} b}$ relates the logarithm of a number $x$ with base $b$ to the logarithm of the same number $x$ with base $k$. The corresponding exponential formula can be obtained by solving for $x$ in the equation:
$\log_{b} x = \frac{\log_{k} x}{\log_{k} b}$
$\log_{b} x \cdot \log_{k} b = \log_{k} x$
$b^{\log_{b} x} = k^{\log_{k} x}$
$x = (k^{\log_{k} x})^{\frac{1}{\log_{b} k}}$
Thus, the exponential form of the logarithmic formula $\log_{b} x = \frac{\log_{k} x}{\log_{k} b}$ is $x = k^{(\frac{1}{\log_{b} k})}$.
I don't feel like this is the correct answer, and I had a long an somewhat interesting conversation with ChatGPT about it's answer that basically ended like: "Ok so that means 1 of the numbers b or k must be equal to x correct?" -> "Yes, that is correct. In the equation $k^{1/(\log_{b} k)} = x$, either $k$ or $b$ must be equal to $x$ in order for the equation to be satisfied."
Here is the equation I believe you are looking for:
$$ b^x \; = \; a^{(\log_a b)x} $$
What follows was extracted and slightly edited from a Fall 1996 handout (handwritten) that I wrote for high school precalculus classes I taught here during 1996-1999. Obviously there are restrictions to the constants and variables below, such as $b \neq 1,$ but to make the derivations less cluttered and thus easier to follow, I'm not bothering to state them here.
Example 1: How can we find a decent decimal approximation of $\log_2 10$ using a calculator that has only log-base-$e$ and log-base-$10$ operations? Write $y = \log_2 10$ in exponential form, then take either $\ln$ or $\log_{10}$ of both sides, then rewrite using a logarithm property, then solve for $y.$ For instance, from $y = \log_2 10$ we get $2^y = 10,$ then taking $\ln$ of both sides gives $\ln (2^y) = \ln 10.$ This last equation can be rewritten as $y \cdot \ln 2 = \ln 10,$ and hence after dividing both sides by $\ln 2$ we get $y = \frac{\ln 10}{\ln 2},$ which is in a form that our calculator can handle.
The same procedure applied to $\log_b x$ gives
$$\log_b x \; = \; \frac{\ln x}{\ln b} \; = \; \left(\frac{1}{\ln b}\right) \cdot \ln x $$
This says that every logarithm operation, to some given base, of $x$ is just a constant (depending only on the base) multiple of $\ln x.$
The same method as in Example 1 (using $a = e^{\ln a})$ can be applied to show more generally (base change for logarithm functions)
$$ \log_b x \; = \; \frac{\log_a x}{\log_a b} \; = \; \left(\frac{1}{\log_a b}\right) \cdot \log_a x $$
Example 2: Notice that using $a = e^{\ln a}$ we get $a^x = (e^{\ln a})^x = e^{x\ln a}.$ Therefore, the exponential function $a^x$ is equal to the function $e^{kx}$ for some number $k$ (which, as you can see, happens to be $\ln a).$ Therefore, the functions $e^{kx}$ (as $k$ varies) include all the functions $a^x$ (as $a$ varies). Moreover, the functions $e^{kx}$ even include all functions of the form $a^{cx}$ for constants $a$ and $c$, since $a^{cx} = (a^x)^c = \left((e^{\ln a})^x\right)^c = e^{(c \ln a)x}$ (i.e. given constants $a$ and $c,$ the "$k$ that works" is $k = c\ln a).$
Exponential growth (occurs with compound interest, unrestricted population growth, radioactive decay, etc.) generally refers to situations in which the independent variable appears linearly as an exponent. From the above, we see that base $e$ (or any other fixed base, say base $10)$ will take care of all bases one might happen to encounter. That is, all exponential growth can be handled using base $e.$
The same method as in Example 2 (using $a = e^{\ln a})$ can be applied to show more generally (base change for exponential functions)
$$ b^x \; = \; a^{(\log_a b)x} $$
Comparing the logarithmic and exponential base change formulas: These formulas are similar in the following way:
(1) If $y = f(x)$ is a logarithmic function, then changing to a new base is equivalent to multiplying $y$ by a certain constant (that depends only on the old and new bases).
(2) If $y = f(x)$ is an exponential function, then changing to a new base is equivalent to multiplying $x$ by a certain constant (that depends only on the old and new bases).