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?
 A: If $t = \exp(x)$, this can be written as $t^{9/29} y = t - 1$, which is a polynomial equation in $t^{1/29}$.  Of course since the degree is greater than $4$, we don't expect a solution involving radicals.  There is, however, a nice series solution in powers of $y$, obtained from the Lagrange Inversion Formula: if $b = 9/29$, 
$$ \eqalign{t &= 1 + y + b y^2 + \dfrac{b(3b-1)}{2!} y^3 + \dfrac{b(4b-1)(4b-2)}{3!} y^4 _+ \ldots\cr
&=   1 + \sum_{n=1}^\infty \left( \prod_{j=0}^{n-2} (nb-j)\right) \frac{y^n}{n!}}$$
A: You will not find any explicit solution if you want $x$ such that $$y=e^{20 x/29} \left(1-e^{-x}\right)$$ Setting $e^x=z$ the equation would write $$y=z^{20/29}(1-\frac 1 z)=\frac{P_{29}(z)}{z^9}$$ with which you would have a lot of fun (just kidding !).
If $y$ is small, you could approximate the function by its Taylor expansion around $x=0$ which will give $$e^{20 x/29} \left(1-e^{-x}\right)=x+\frac{11 x^2}{58}+\frac{301 x^3}{5046}+O\left(x^4\right)$$ and using more or less terms, you could find an approximation of $x$ as a function of $y$. But this works for a very limited range of $y$.
For the most general case, the only way is to consider a method such as Newton which, starting at a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For your case $$f(x)=e^{20 x/29} \left(1-e^{-x}\right)-y$$ $$f'(x)=\frac{1}{29} e^{-9 x/29} \left(20 e^x+9\right)$$ For illustration purposes, let us consider two cases.
First case : $y=12345$. A look at the plot of the function shows a solution around $x=15$. So, let us apply the method using $x_0=15$. The successive iterates will then be 
$$\left(
\begin{array}{cc}
n & x_n \\
 1 & 14.12564931 \\
 2 & 13.72770282 \\
 3 & 13.66199628 \\
 4 & 13.66046179 \\
 5 & 13.66046098 
\end{array}
\right)$$
Second case : $y=-12345$. A look at the plot of the function shows a solution around $x=-30$. So, let us apply the method using $x_0=-30$. The successive iterates will then be 
$$\left(
\begin{array}{cc}
n & x_n \\
 1 & -30.37705418 \\
 2 & -30.35664112 \\
 3 & -30.35657618
\end{array}
\right)$$
A: 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.
