Given a real $c$ such that $1 < c$, is there any known and direct lower bound, other than $0$, for $(\ln c)$, i.e., $A < \ln c$?
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1Do you want a lower bound with any special properties? What makes a lower bound "direct"? – hardmath Nov 27 '13 at 19:00
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just a lower bound $A$ such that $0 < A$. Thanks. – Mai09el Nov 28 '13 at 00:22
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For all $x \ge 1$ we have $\log x \ge \frac{x-1}{x+1}$, with equality only at $x=1$.
Note that the derivative of $\log x$ is $1/x$, and the derivative of the rational function, which we can rewrite as $1 - 2(x+1)^{-1}$, is $2(x+1)^{-2}$:
$$ x \ge 1 \implies \frac{1}{x} \gt \frac{2}{(x+1)^2} $$
hardmath
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3The lower bound can be "sharpened" to $2\frac{x-1}{x+1}$, giving an equal slope "osculation" at $x=1$, but the Question does not require this. – hardmath Nov 28 '13 at 01:54
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Since $\lim_{x\to 1}\ln x=0$, there can be no positive lower bound for the natural logarithms of numbers greater than $1$.
More explicitly, given $A>0$ you can always find $c>1$ such that $0<\ln c<A$: just take $c=e^{A/2}$.
egreg
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While no constant $A$ is valid (as your argument shows), the OP is willing to take a rational function as the lower bound. – hardmath Nov 28 '13 at 02:15
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Yes. $\ln c \geq 0$ for $c \geq 1$, with equality only at $c=1$.
apt1002
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Except that the question said "other than $0$". I'd guess he means some easily computed lower bound, for example maybe a rational function. – Michael Hardy Nov 27 '13 at 19:20
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As stated in the other answers, there is no constant $A \gt 0$ as a lower limit for $\ln c$, $c \gt 1$. However, you can use the approximation to $ln$ at 1, as given on the Wikipedia page, which starts $ln (1+x) \approx x - \dfrac{x^2}2\ldots$
half-integer fan
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Since we were discussing bounds A "close to" 0, I assumed we were only interested in a bounding function for $x << 1$. – half-integer fan Nov 28 '13 at 02:29