Infinite Logging

$\begin{aligned} 9 & = \sum_{n=0}^\infty \frac {\lg x^n}{\lg^n x} = \sum_{n=0}^\infty \frac {n \lg x}{\lg^n x} = \sum_{n=1}^\infty \frac n{(\lg x)^{n-1}} & \small \blue{\text{Let }u = \frac 1{\lg x}} \\ & = \sum_{n=1}^\infty \frac nu^{n-1} = \sum_{n=1}^\infty \frac d{du} u^n = \frac d{du} \sum_{n=1}^\infty u^n & \small \blue{\text{For }u< 1} \\ & = \frac d{du} \left(\frac 1{1-u}\right) = \frac 1{(1-u)^2} \end{aligned}$

Therefore,

$\begin{aligned} \frac 1{(1-u)^2} & = 9 \\ 1-u & = \frac 13 \\ \implies u & = \frac 23 & \small \blue{\text{Note that }u = \frac 1{\lg x}} \\ \lg x & = \frac 32 \\ \implies x & = 10^\frac 32 = \boxed{10\sqrt{10}} \end{aligned}$

$\begin{aligned} \displaystyle\sum_{n=0}^\infin\frac{\lg\left(x^n\right)}{\lg^n\left(x\right)}&=9\\ \frac{\lg\left(x^1\right)}{\lg^1\left(x\right)}+\frac{\lg\left(x^2\right)}{\lg^2\left(x\right)}+\frac{\lg\left(x^3\right)}{\lg^3\left(x\right)}+\cdots&=9\\ \frac{1\lg\left(x\right)}{\lg\left(x\right)}+\frac{2\lg\left(x\right)}{\lg^2\left(x\right)}+\frac{3\lg\left(x\right)}{\lg^3\left(x\right)}+\cdots&=9\\ 1+\frac{2}{\lg\left(x\right)}+\frac{3}{\lg^2\left(x\right)}+\cdots&=9\\ \frac{2}{\lg\left(x\right)}+\frac{3}{\lg^2\left(x\right)}+\frac{4}{\lg^3\left(x\right)}+\cdots&=8\\ \text{Let }u&=\lg\left(x\right)\\ \frac{2}{u}+\frac{3}{u^2}+\frac{4}{u^3}+\cdots&=8\\ 2+\frac{3}{u}+\frac{4}{u^2}+\cdots&=8u\\ 2+\left(\frac{2}{u}+\frac{3}{u^2}+\frac{4}{u^3}+\cdots\right)+\left(\frac{1}{u}+\frac{1}{u^2}+\frac{1}{u^3}+\cdots\right)&=8u\\ 2+8+\sum_{m=1}^{\infin}\frac{1}{u^m}&=8u\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=8u-10 \end{aligned}$

$\begin{aligned} \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}+\frac{1}{u^2}+\frac{1}{u^3}+\cdots\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}\left(1+\frac{1}{u}+\frac{1}{u^2}+\cdots\right)\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}\left(1+\sum_{m=1}^{\infin}\frac{1}{u^m}\right)\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}+\frac{1}{u}\sum_{m=1}^{\infin}\frac{1}{u^m}\\ \sum_{m=1}^{\infin}\frac{1}{u^m}-\frac{1}{u}\sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}\\ \left(1-\frac{1}{u}\right)\sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u}\frac{1}{1-\frac{1}{u}}\\ \sum_{m=1}^{\infin}\frac{1}{u^m}&=\frac{1}{u-1} \end{aligned}$

$\begin{aligned} \left|\frac{1}{u}\right|&<1\\ -1<\frac{1}{u}&<1\\ u<-1&\text{ or }u>1\\ \end{aligned}$

$\begin{aligned} \sum_{m=1}^{\infin}\frac{1}{u^m}&=8u-10\\ \frac{1}{u-1}&=8u-10\\ 1&=\left(8u-10\right)\left(u-1\right)\\ 1&=8u^2-8u-10u+10\\ 8u^2-18u+9&=0\\ (2u-3)(4u-3)&=0\\ \end{aligned}$

$\begin{matrix} \begin{aligned} 2u-3&=0\\ 2u&=3\\ u&=\frac32\\ \lg\left(x\right)&=\frac32\\ x&=10^\frac32\\ x&=10^{\frac22+\frac12}\\ x&=\left(10^1\right)\left(10^\frac12\right)\\ x&=\boxed{10\sqrt{10}} \end{aligned} &\text{ or }& \begin{aligned} 4u-3&=0\\ 4u&=3\\ u&=\frac34\,(\text{Rej.}) \end{aligned} \end{matrix}$

Let $log_{10}x=y$ . Then $1+\dfrac{2}{y}+\dfrac{3}{y^2}+\dfrac{4}{y^3}+....=9$ . Or $(\dfrac{y}{y-1})^2=9$ , or $y=log_{10}x=\dfrac{3}{2}$ or $y=log_{10}x=\dfrac{3}{4}$ . Since the sum converges to $9$ , $y$ must be greater than $1$ . So $x=10^{3/2}=\boxed {10\sqrt {10}}$ .