# Logarithmic inequalities

Full disclosure: This is a homework problem, but my question is regarding a concept that came about during solving the problem, not the actual solution to the problem.

Problem: Rewrite as geometric power series (might not be correct formulation of problem, I'm still learning :] ) $$\sum_{n=0}^{\infty} (\ln x)^n$$

Formulating this as a geometric series with $a_1=1$ and $r=\ln x$, it converges to $$S=\frac{1}{1-\ln x}$$ if $|\ln x| < 1$.

I rewrite the above inequality to get rid of the absolute value as $-1 < \ln x < 1$.

Now, when I exponentiate (to undo the log), I get $$e^{-1} < x < e^1$$

The question: If the interval contained negative numbers, would I truncate it just to contain positive numbers as the argument of a log cannot be negative?

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Do you mean your final interval? That can't happen, since a positive number raised to any power (positive or negative) is positive. (After all, that's why the argument of a log can't be negative.)

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now that you put it that way the base can't be negative so exponentiation will never beget a negative value. Makes sense thanks :) –  Jane Doe Feb 10 '14 at 0:39
@JaneDoe: With just a little work, I think you can show this formula is valid even for complex values $z=re^{i\theta}$ as long as $e^{-1}<r<e$ and $|\theta| < \sqrt{1 - (\ln r)^2}$. –  MPW Feb 10 '14 at 1:08