solve equation with sum $\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...=2$

Question

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.

Answer 1

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 $x = e^3$.

Answer 2

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:

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

$$\ln x=3$$

$$x=e^3$$

Answer 3

$$\begin{align} 2&=\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+\cdots\\ &=\ln x-\ln x^{\frac 12}+\ln x^{\frac 14}-\ln x^{\frac 18}+\cdots \\ &=\ln\left(\frac{x\cdot x^{\frac 14}\cdot x^{\frac 1{16}}\cdots}{\;\;x^\frac 12\cdot x^\frac 18\cdot x^{\frac 1{32}}\cdots }\right)\\ &=\ln \left(x^{\frac 12}\cdot x^{\frac 18}\cdot x^{\frac 1{32}}\cdots\right)\\ &=\ln \left(x^{\frac 12+\frac 18+\frac 1{32}+\cdots}\right)\\ &=\left(\frac 12+\frac 18+\frac 1{32}+\cdots\right)\ln x\\ &=\frac {\frac 12}{1-\frac 14}\ln x\\ &=\frac 23 \ln x\\ 3&=\ln x\\ x&=e^3\quad\blacksquare \end{align}$$