I am struggling to find the inverse of the following function $$f(x) = \frac{10}{3}\exp\big(-0.06x\big)\log\bigg(\frac{1}{5}\big(2x + 3\big)\bigg).$$ I noticed that this function is not one-to-one, so I restricted its domain to the interval $[1, 9.5]$. I am aware that there may not be an elementary answer, so while I may prefer an elementary answer, I would be very glad to get an answer in the form of something like a series.
Thank you in advance!
I can almost guarantee that this will not have an elementary inverse. If your need is simply to calculate values, then I suggest a root-finding technique. You have this tagged as pre-calculus, but I'm going to bend that a bit and suggest Newton's method. Though setting up the recurrence for Newton's method requires a derivative, that only needs performed once, and then the rest is just algebra.
Suppose you want to find $f^{-1}(a)$ for some $a$. Define $g(x) = f(x) - a$. The problem is now one of finding the root of $g$. Newton's method sets up the recurrence
$$x_{n+1} = x_n - \dfrac{g(x_n)}{g'(x_n)} = x_n - \dfrac{f(x) - a}{f'(x)}$$
Which hopefully will converge to the root of $g$, which is $f^{-1}(a)$. For your function, if you start with $x_0 = 1$, it will converge very quickly for $a < 2.5$. As $a$ gets close to its maximum (a little under $3$), convergence will slow, but the sequence should still converge instead of blowing up, as occasionally happens with Newton's method.
$$f'(x) = \frac{10}3e^{-0.06x}\left[(-0.06)\log\left(\dfrac{2x+3}5\right) + \dfrac2{2x+3}\right]$$
So the method becomes (with some simplification) $$\begin{align}x_0 &= 1\\x_{n+1} &= x_n - \dfrac{\dfrac{50}3\log\left(\dfrac{2x_n + 3}5\right) - 5a\,e^{0.06x_n}}{\dfrac {100}{6x_n+9} - \log\left(\dfrac{2x_n+3}5\right)}\end{align}$$ which will converge to $f^{-1}(a)$ for $a \in (-\infty, 2.7949)$, giving values in $\left(-1.5, 9.6483\right)$.
