Interval of the solutions to $\log_{1/2}\log_2(\frac{1+2x}{1+x})>0$ is?

Question

I consistently get $x>-1$ but that doesn't fit the possible solutions I've got.

First step I do is state that $\log_2(\frac{1+2x}{1+x})<1$

Then express the $1$ as $\log_22$ and so on.

What am I doing wrong?

Answer 1

You need to have the followings as well : $$\log_2\left(\frac{1+2x}{1+x}\right)\gt 0\ \ \ \ \text{and}\ \ \ \ \frac{1+2x}{1+x}\gt 0,$$ i.e. $$\frac{1+2x}{1+x}\gt 1.$$

Hence, the answer will be $$x\gt -1\ \ \ \ \text{and}\ \ \ \ \frac{1+2x}{1+x}\gt 1,$$ i.e. $$x\gt 0.$$