Logarithm in an inequality: is it solvable?

Question

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? $$ \left \{ \begin{aligned} \ln{(- \varepsilon )}&< \ln{2^{\frac{1}{x}} }\\ \ln{( \varepsilon )}&> \ln{2^{\frac{1}{x}} } \end{aligned} \right. $$

Answer 1

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}