Isolate x in the equation I'm trying to isolate $x$ from the next equation. Some ideas?
\begin{equation}
k_{1} = k_{2}\sum_{i=1}^{x}\binom{x-1}{i-1}\frac{1}{i^{k_{3}}},
\end{equation}
where $k_{1}$, $k_{2}$ and $k_{3}$ are constants. If not possible to do it exactly, an approximation can also be used.
Thank you in advance by your help!
 A: Forgive me as this is not entirely an answer (I do not have comment privileges) but by observing only 
$$f(c)=\sum_{i=1}^x\binom{x-1}{i-1} \dfrac{1}{i^c}:c\in\mathbb{Z}$$
we can make some observations. Of course, in your case, $k_3\in\mathbb{R}$
I have listed some observations with various integer values below:
\begin{align}
&\quad\!\!\!\cdot\\
&\quad\!\!\!\cdot\\
&\quad\!\!\!\cdot\\
f(-3)&=(x+1)(x^2+5x-2)2^{x-4}\\
f(-2)&=x(x+3)2^{x-3}\\
f(-1)&=(x+1)2^{x-2}\\
f(0)&=2^{x-1}\\
f(1)&=\frac{2^{x-1}}{x}\\
f(2)&=_3\!\!F_2(1,1,1-x;2,2;-1)\\
f(3)&=_4\!\!F_3(1,1,1,1-x;2,2,2;-1)\\
&\quad\!\!\!\cdot\\
&\quad\!\!\!\cdot\\
&\quad\!\!\!\cdot\\
\end{align}
where $_pF_q(a_1,\cdots,a_p;b_1,\cdots,b_q;x)$ is the generalized hypergeometric function given by
 $$_pF_q(a_1,\cdots,a_p;b_1,\cdots,b_q;x)=\sum_{k=0}^\infty \dfrac{(a_1)_k\cdots(a_p)_kx^k}{(b_1)_k\cdots(b_q)_kk!}$$
We do notice some behavior with the function $f(c)$ as the polynomial "coefficient" decreases its degree by 1 with on increase of $c$ by $1$ and the exponent power increases by $1$ up until $f(0)$.
In general for $c=k_3\in\mathbb{Z^{>2}}$,
$$f(c)=_{c+1}F_{c}(\underbrace{1,\cdots,1}_{c+1 \quad\!\!\! \text{times}},1-x;\underbrace{2,\cdots 2}_{c \quad\!\!\! \text{times}};-1)$$
