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? $$ \begin{cases} \ln (- \varepsilon ) < \ln 2^{\frac{1}{x}}\\ \ln( \varepsilon ) > \ln 2^{\frac{1}{x}} \end{cases} $$
Logarithm in an inequality: is it solvable?
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1First, isn't a power of a number always positive? (Unless you are working with complex powers, but I don't think that's the case here). Second, can you take the logarithm of a negative number? (Again, no complex numbers assumed). – 2012-09-30