How do solve an sum of numbers to a common power? How would solve for $a$ in this equation without using an approximation  ?
is it possible?
where $x>0$ and $0<a<\infty$
$x=\Sigma _{i=1}^{n} i^a$
for example $120=\Sigma _{i=1}^{6} i^a$ what is $a$ in this equation?
 A: The sum you give yields the Harmonic number, $H_n^{(-a)}$.  This can be solved, in Mathematica for instance:
Solve[k == HarmonicNumber[n, -a], a][[1]]

When you plug in $k=120$ and $n=6$ as required by the problem example one finds:
$a = 2.17926794327541855469750380268$
As a check:
$\sum\limits_{i=1}^6 i^a = 119.9999$ (close enough!).
A: If you want solve for $a$ the equation $$x=\sum_{i=1}^{n} i^a=H_n^{(-a)}$$ you will need a numerical method (as already said in answers) and, for that, you would need some initial estimate.
To get such an estimate, you could consider the asymptotics
$$H_n^{(-a)}=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$ and set the problem as
$$x \approx \frac{n^{a+1}}{a+1}$$ for which the solution is given in terms of Lambert function. This will be $$a=-1-\frac{1}{\log (n)}\,W_{-1}\left(-\frac{\log (n)}{k}\right)$$ Since the argument of the function is supposed to be small, you could use for its evaluation
$$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ (as given in the linked page).
For your specific example $(n=6,x=120)$, this would give $a\approx 2.34602$  which is not too bad for an estimate.
Now to polish the root, consider that you look for the zero of function
$$f(a)=\sum_{i=1}^{n} i^a-x$$ $$f'(a)=\sum_{i=1}^{n}i^a \log(i)$$ and use Newton method which, starting from the guess $a_0$, will update it according to
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$$ Applied to the example, this will give as iterates
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 2.34602 \\
 1 & 2.19989 \\
 2 & 2.17961 \\
 3 & 2.17927 
\end{array}
\right)$$
If you do not want to use Lambert function, you could consider that $$\sum_{i=1}^{n} i^a <\sum_{i=1}^{n} n^a=n^{a+1}\implies a_0=\frac{\log (x)}{\log (n)}-1$$ This would be simpler but will require more iterations as shown below
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 1.67195 \\
 1 & 2.45312 \\
 2 & 2.23205 \\
 3 & 2.18146 \\
 4 & 2.17927
\end{array}
\right)$$
Edit
Since function $f(a)$ varies very fast with $a$, it could be better to make it more linear considering instead
$$g(a)=\log\left(\sum_{i=1}^{n} i^a \right)-\log(x)$$
$$g'(a)=\frac{\sum_{i=1}^{n} i^a \log(i) }{\sum_{i=1}^{n} i^a }$$
Using the simple estimate, for the worked example, the iterates would be
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 1.67195 \\
 1 & 2.18953 \\
 2 & 2.17927
\end{array}
\right)$$
Just for the fun, let us use $(n=20,x=123456789)$; for this case, the iterates would be
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 5.21931 \\
 1 & 5.80609 \\
 2 & 5.80465
\end{array}
\right)$$
A: There is not a general closed form but we can use integral estimation
$$\sum_{i=1}^{6} i^a\approx \int_{0.5}^{6.5}x^adx=\left[\frac{x^{a+1}}{a+1}\right]_{0.5}^{6.5}=\frac{6.5^{a+1}}{a+1}-\frac{0.5^{a+1}}{a+1}=120 \implies a\approx 2.175$$
where the value for $a$ needs to be evaluated by numerical methods and by a direct check we obtain
$$\sum_{i=1}^{6} i^{2.175}\approx 119.21$$
A: If you set $a=\log t$, you can write the question as
$$x=\sum_{k=1}^n k^a=\sum_{k=1}^n t^{\log k}$$
where the RHS is a generalized polynomial (with irrational powers). So this is even less solvable than a polynomial and you can't avoid numerical methods.
The function (of $a$) is strictly growing, comes from $0$ in the negatives, equals $n$ for $a=0$, then tends to $n^a$ asymptotically. A reasonable starting value for Newton's iterations is given by $\log_n x$.
