As it is quite easy to calculate integer solutions for $\;^hb = b^h $ the question arose on how to find a solution for real heights. One idea was of predicting the height by finding a formula approximating it. For more on that specific equation see Might results which show the same result for tetration as for exponentiation be of any use (like in the range from 2 to e^(1/e))? . However this just gives probably just one solution for one base between 2 and $e^{1/e} $ and so I thought of changing it to $\;^hb = b^h + c $ which can just be approximated or simplified to $\; ^hb \approx c$ .
I tried with different constants and decided to use Newton's divided differences interpolation polynomial method to get a good approximation for the desired interval as predicting the whole series of solutions from $2$ to infinity with some self-made formula seemed to be hopeless as it would never be exact enough -- especially for lower heights. It turned out it just needed to be pretty good in the range from $3$ to $4$ for base $e$ and $4$ to $5$ in for base $2$. I tried with different numbers of polynomials and different constants.
Of interested was just the non-integer or fractional value due to the unique property of tetration where $\; ({^{b}a})^{(^{c}a)} = ({^{(c+1)}a}) ^ {({^{(b-1)}a})}$. From this can be concluded that from a tetrational height $x$ between 0 and 1 all other heights n+x can be calculated where $n$ is any integer greater than $0$.
Results
I used a constant c of $\;^4(2.1) $ ~ 84899120142.81 to predict the calculated value for $\;^4b = c $ and the best prediction was around 2.15 instead of the calculated 2.1000003814697266. Then I supposed it would behave like $2^x $ and used multiplication to get more values and then I used the Newton polynomial method to get an approximation and the result was the formula $1 + 0.6856427806944169 x + 0.22005174698143642 x^2 + 0.07447687052884291x^3$.
In a similar way I used different constants to get solutions for $\;^3b = c $ and from the results I used Newton polynomials to get a formula or function which approximates the exponential function which pretty good approximates $\;^xe $ for $0 > x < 1 $. The result was $1 + 1.0432362881329733 x + 0.5910113778264401 x^2$.
Please take a look at https://github.com/MatthiasLiszt/tetrationalPlayground/ to see graphs and code.
