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.
Let

$\begin{array}\\
f(x)
&amp;=\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...\\
&amp;=\sum_{n=0}^{\infty} (-1)^n\ln(\sqrt[2^n]{x})\\
&amp;=\sum_{n=0}^{\infty} (-1)^n\frac{\ln(x)}{2^n}\\
&amp;=\ln(x)\sum_{n=0}^{\infty} (-1)^n\frac{1}{2^n}\\
&amp;=\ln(x)\frac1{1+1/2}\\
&amp;=\ln(x)\frac{2}{3}\\
\end{array}
$

So,
if $f(x) = 2$,
$\ln(x) = 3$
and
$x = e^3$.
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:

$$\frac 23\ln x=2$$

$$\ln x=3$$

$$x=e^3$$