Logarithm in an inequality: is it solvable? Can anyone help me understand what happens to the following inequality once I apply a logarithm to all three parts?
$$
- \varepsilon < 2^{\frac{1}{x}} < \varepsilon \longrightarrow \ln(- \varepsilon )< \ln 2^{\frac{1}{x}} < \ln\varepsilon
$$
where $\epsilon > 0$. Can I rewrite it this way? Are they both correct and solvable?
$$
\begin{cases}
 \ln (- \varepsilon ) < \ln 2^{\frac{1}{x}}\\
 \ln( \varepsilon ) > \ln 2^{\frac{1}{x}}
\end{cases}
$$
 A: If $x$ is a real number then $2^{1/x}$ is necessarily positive, so the set of values of $x$ for which $-\varepsilon<2^{1/x}<\varepsilon$ is the same as the set of values of $x$ for which  $2^{1/x}<\varepsilon$.  If $x$ is not a real number, then there's the problem of what the inequalities mean.
You can't take $\ln$ of a negative number, and if you could, there's the big fat question of whether it would preserve inequalities.
Certainly it preserves inequalities among positive numbers, i.e. $\ln$ is an increasing function.
So you have $2^{1/x}<\varepsilon$; hence $\ln(2^{1/x})<\ln\varepsilon$.
If you don't know that that implies $\frac1x\ln 2<\ln\varepsilon$ then you really need to work on your understanding of logarithms before getting into problems like this.
From there, and the fact that $\ln2$ is positive, it's easy to deduce that $x>\dfrac{\ln2}{\ln\varepsilon}= \dfrac{1}{\log_2\varepsilon}$.
Maybe using base-$2$ logarithms is a bit quicker:
\begin{align}
2^{1/x} & < \varepsilon \\[6pt]
\frac1x & < \log_2\varepsilon \\[6pt]
x & > \frac{1}{\log_2\varepsilon}
\end{align}
