can we express $x=f(y)$, if $y=\exp(x/1.45)(1-\exp(-x))$? Given $y=\exp(x/1.45)(1-\exp(-x))$, I need to express $x = f(y)$.
I tried taking log of both side but didnt get any fruitful expression.
Is it possible? If yes, how?

Here are the details of the Lagrange Inversion computation. We have
$$y = \exp(bx)(1-\exp(-x)).$$
where $b=20/29.$ Putting $\exp(x) = t$ so that $x = \log t$ we get
$$y = t^b (1-1/t).$$
We seek to compute
$$[y^n] t(y) =
\frac{1}{2\pi i}
\int_{|y|=\epsilon} \frac{1}{y^{n+1}} t(y) dy.$$
Using the definition we put with $w$ being complex
$$y = w^b (1-1/w) = w^{b-1} (w-1)$$
so that
$$dy = (b w^{b-1} - (b-1) w^{b-2}) \; dw.$$
We get for the integral
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{1}{w^{(b-1)(n+1)}}
\frac{1}{(w-1)^{n+1}}
\\ \times w \times
w^{b-2} (bw - (b-1))
\; dw.$$
This has two pieces, the first is
$$\frac{b}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{1}{w^{(b-1)n-1}}
\frac{1}{(w-1)^{n+1}}
\; dw
\\ = \frac{b}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{1}{(1+(w-1))^{(b-1)n-1}}
\frac{1}{(w-1)^{n+1}}
\; dw
$$
and the second is
$$\frac{b-1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{1}{w^{(b-1)n}}
\frac{1}{(w-1)^{n+1}}
\; dw
\\ = \frac{b-1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{1}{(1+(w-1))^{(b-1)n}}
\frac{1}{(w-1)^{n+1}}
\; dw.$$
Extracting coefficients yields
$$(-1)^n
\left(b {(b-1)n-2+n\choose n}
- (b-1) {(b-1)n-1+n\choose n}\right)
\\ = (-1)^n
\left(b {bn-2\choose n}
- (b-1) {bn-1\choose n}\right).$$
This is one when $n=0.$ Continuing we have
$$\frac{(-1)^n}{n!}
(b (bn-2)^{\underline{n}} - (b-1) (bn-1)^{\underline{n}})
\\ = \frac{(-1)^n}{n!} (bn-1)^{\underline{n}}
\left(b \frac{bn-(n+1)}{bn-1} - (b-1)\right)
\\ = \frac{(-1)^{n+1}}{n!} (bn-1)^{\underline{n}}
\frac{1}{bn-1}
\\ = \frac{(-1)^{n+1}}{n!} (bn-2)^{\underline{n-1}}.$$
We thus obtain
$$t = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1}
\prod_{j=0}^{n-2} (nb-2-j)
\\ = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1}
\prod_{j=2}^{n} (nb-j).$$
Remark.  Observe that  the  accepted answer  and  this post  solve
slightly different equations which both yield the correct result.
