solve equation with sum $\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...=2$ How to solve this? Any advice?
$$\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...=2$$
Next step I do this
$\sum\limits_{n=0}^\mathbb{\infty}(-1)^n\ln(x^{\frac{1}{2^n}}) = 2$
But I don't know next step. Thanks.
 A: Let
$\begin{array}\\
f(x)
&=\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...\\
&=\sum_{n=0}^{\infty} (-1)^n\ln(\sqrt[2^n]{x})\\
&=\sum_{n=0}^{\infty} (-1)^n\frac{\ln(x)}{2^n}\\
&=\ln(x)\sum_{n=0}^{\infty} (-1)^n\frac{1}{2^n}\\
&=\ln(x)\frac1{1+1/2}\\
&=\ln(x)\frac{2}{3}\\
\end{array}
$
So,
if $f(x) = 2$,
$\ln(x) = 3$
and
$x = e^3$.
A: $$\begin{align}
2&=\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+\cdots\\
&=\ln x-\ln x^{\frac 12}+\ln x^{\frac 14}-\ln x^{\frac 18}+\cdots \\
&=\ln\left(\frac{x\cdot x^{\frac 14}\cdot x^{\frac 1{16}}\cdots}{\;\;x^\frac 12\cdot x^\frac 18\cdot x^{\frac 1{32}}\cdots
}\right)\\
&=\ln \left(x^{\frac 12}\cdot x^{\frac 18}\cdot x^{\frac 1{32}}\cdots\right)\\
&=\ln \left(x^{\frac 12+\frac 18+\frac 1{32}+\cdots}\right)\\
&=\left(\frac 12+\frac 18+\frac 1{32}+\cdots\right)\ln x\\
&=\frac {\frac 12}{1-\frac 14}\ln x\\
&=\frac 23 \ln x\\
3&=\ln x\\
x&=e^3\quad\blacksquare
\end{align}$$
A: For the property of logarithms you can rewrite your sum as:
$$\sum_{n=0}^{\infty}\left(-\frac 12\right)^n\ln x=\ln x-\frac 12\ln x+\frac 14\ln x-\dots=2$$
Now remember the formula for the geometric series:
$$\frac 1{1-z}=\sum_{n=0}^{\infty}z^n=1+z+z^2+z^3+...$$
You can easily see from there that:
$$1-\frac 12+\frac 14-\frac 18+\dots=\frac 23$$
So you have that:
$$\sum_{n=0}^{\infty}\left(-\frac 12\right)^n\ln x=\ln x\sum_{n=0}^{\infty}\left(-\frac 12\right)^n=\frac 23\ln x$$
Now your equation is:
$$\frac 23\ln x=2$$
$$\ln x=3$$
$$x=e^3$$
