Solve infinite series equation with logarithmic terms. Solve logarithmic equation: $$\frac{\log x^2}{\log^{2}x}+\frac{\log x^3}{\log^{3}x}+\cdots+\frac{\log x^k}{\log^{k}x}+\cdots=8$$
here $\log$ is assumed to have base $10$.
So far I managed to rewrite it as: $$\sum_{n=2}^{\infty} \frac{n}{\log^{n-1}{x}}=8 \iff \sum_{n=2}^{\infty} n\log_{x}^{n-1}{10}=8$$
But I don't know how to find the infinite sum.
 A: We know that:
$$\sum_{n=1}^\infty nf^{n-1}(x)=\frac 1{(1-f(x))^2}$$
Now you say that you have rearranged your equation as:
$$\sum_{n=2}^\infty n\log_x^{n-1}10=8$$
So you now have to solve the equation:
$$\frac 1{(1-\log_x10)^2}-1=8$$
Rearranging the terms leads us to:
$$\log_x10=\frac 23$$
Which solution is: $10\sqrt {10}$
A: Observe: Since
$$
\sum_{n=2}^\infty\frac{n}{\log^{n-1}(x)}=8,
$$
we know that by multiplying by $\frac{1}{\log x}$, we have
$$
\sum_{n=2}^\infty\frac{n}{\log^{n}(x)}=\frac{8}{\log x},
$$
Therefore,
$$
\sum_{n=2}^\infty\frac{n}{\log^{n-1}(x)}-\sum_{n=2}^\infty\frac{n}{\log^{n}(x)}=8-\frac{8}{\log x}.
$$
Next, by reindexing the second sum, we have
$$
\sum_{n=2}^\infty\frac{n}{\log^{n-1}(x)}-\sum_{n=3}^\infty\frac{n-1}{\log^{n-1}(x)}=8-\frac{8}{\log x}.
$$
Canceling in the sums we have
$$
\frac{2}{\log x}+\sum_{n=3}^\infty\frac{1}{\log^{n-1}(x)}=8-\frac{8}{\log x}.
$$
The remaining sum is a geometric series (and assuming $|\log x|>1$), we have
$$
\frac{2}{\log x}+\frac{1}{\log^2 x}\sum_{n=0}^\infty\frac{1}{\log^{n}(x)}=8-\frac{8}{\log x}.
$$
Simplifying the geometric series, we have
$$
\frac{2}{\log x}+\frac{1}{\log^2 x}\frac{\log x}{\log x-1}=8-\frac{8}{\log x}.
$$
Writing $y=\log x$, we have
$$
\frac{2}{y}+\frac{1}{y}\frac{1}{y-1}=8-\frac{8}{y}.
$$
Clearing fractions, we have
$$
2(y-1)+1=8y(y-1)-8(y-1).
$$
Simplifying, we know that
$$
0=8y^2-18y+9.
$$
Applying the quadratic formula, we have
$$
y=\frac{3}{4},\frac{3}{2}.
$$
Since, we assumed earlier that $|y|>1$, it follows that 
$$
\log x=\frac{3}{2}
$$
or that
$$
x=10^{3/2}.
$$
This answer checks with wolfram alpha.
A: Let $x=10^{y}$. Then we have
$$\sum_{n=2}^\infty ny^{1-n}$$
Setting $t=y^{-1}$ we get
$$\sum_{n=2}^\infty nt^{n-1}={1\over (1-t)^2}-1.$$
We can see this formula is true because when multiplying two power series, $\sum a_nx^n, \sum_nb_nx^n$ we get the coefficient of $x^n$ is $\sum_{k+j=n}a_kb_j$ and
$$\sum_{k+j=n}1\cdot 1=n$$
since there are $n$ ways to write $n=k+j$, namely $1+(n-1), 2+(n-2),\ldots (n-1)+1$.
Setting this equal to $8$, we get $(1-t)^2={1\over 9}$, i.e.
$$1-t=\pm {1\over 3}.$$
Then this means
$$t={3\pm 1\over 3}\iff y= {3\over 3\pm 1}$$
$|y|>1$ for this to converge, so $y=3/2$.
So $x=10^{3/2}=10\sqrt{10}$.
