Solve $\log_2( \log_{1/2}(\vert x \vert - 1 )) > 0 $

Question

I am new to logarithmic inequalities. How to find the solution set for this inequation:

$$\log_2( \log_{1/2}(\vert x \vert - 1 )) > 0 $$

I tried by taking two cases:

$$ \log_2( \log_{1/2}( x - 1 )) > 0$$ and $$ \log_2( \log_{1/2}(- x - 1 )) > 0$$

Then by solving them, I got $ x \in \left(\frac{1}{2},\frac{3}{2}\right) $

Sorry, it was a silly calculative error

I got $ x \in \left(\frac{-3}{2},\frac{3}{2}\right) $

Is this the correct way? Can anyone provide a good or a robust way?

Answer 1

We can use the equality that $\log_ab=c\Longleftrightarrow a^c=b$

Therefore, we have \begin{align}\log_2(\log_{1/2}(|x|−1))&>0\\ \log_{1/2}(|x|−1)&>2^0\\ \log_{1/2}(|x|−1)&>1\\ 0<|x|−1&<\left(\frac12\right)^1\\ 0<|x|−1&<\frac12\end{align}

We have to flip the sign when raising $\frac 12$ to a power, as larger powers mean smaller values. We must also ensure that the inside of the logarithm is always gresater than $0$ as we cannot take a logarithm of a negative number.

Now we can solve this as we would with a normal absolute value question. We split it into two halves and solve each half.

When $|x|=x$:

\begin{align}0&<x-1<\frac 12\\ 1&<x<\frac 12 + 1\\ 1&<x<\frac 32\end{align}

When $|x|=-x$:

\begin{align}0&<-x-1<\frac 12\\ 1&<-x<\frac 12+1\\ 1&<-x<\frac 32\\ -1&>x>-\frac32\end{align}

So, we can conclude that $$x\in\left(-\frac32,-1\right)\cup\left(1,\frac 32\right)$$

Answer 2

Properties used here:

1. If $a > 1$, then $\log_a M > b \iff M > a^b$
2. If $0 < a < 1$, then $\log_a N > b \iff N < a^b$

Applying the first property, we see that $\log_2M > 0$ means $M > 2^0 = 1$. So you need to have $\log_{1/2}(|x| - 1) > 1$.

Now apply the second property: $\log_{1/2} N > 1$ means $N < (1/2)^1 = 1/2$. So now you need $|x| - 1 < 1/2$.

Recall that anything you take a logarithm of (regardless of the valid base) must be positive. So you must also have $|x| - 1 > 0$. (This domain restriction also implies that $\log_{1/2}(|x| - 1)$ must be positive, but this is already enforced by the earlier fact that we must have $\log_{1/2}(|x|-1) > 1$.)