how can we prove ln(1/a) = -ln(a)

by using fundermental theory of calculus?

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- May 6th 2010, 03:02 AM #1

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- May 6th 2010, 04:04 AM #2

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- May 6th 2010, 04:05 AM #3

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Let $\displaystyle a > 0$, then $\displaystyle \ln a = \int\limits^a_1 \frac{dt}{t}$; make change of variable: $\displaystyle u = \frac{1}{t}\Longrightarrow du = -\frac{dt}{t^2}\Longrightarrow$ $\displaystyle dt = -\frac{du}{u^2}$ , and the limits: $\displaystyle t = 1 \Longrightarrow u = 1\,,\,t=a\Longrightarrow u=\frac{1}{a}$ , so :

$\displaystyle \ln a = \int\limits^a_1 \frac{dt}{t}=-\int\limits_1^{1/a}\frac{du}{u}=-\ln(1/a)$

Tonio

- May 6th 2010, 03:43 PM #4

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Me too it is a similar problem that brought me to this site. Essentially ShaXar wants to find the intersection of the curves exp(x) and 1/x or the curves 1/x and ln(x), i.e. where they intersect but it is not so simple.

How on earth can one solve such an ordinary and seemingly so simple a problem, anyone ???

- May 7th 2010, 02:09 AM #5

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If $\displaystyle e^x= \frac{1}{x}$, then $\displaystyle xe^x= 1$ so x= W(1) where W(x) is the "Lambert W function", Lambert W function - Wikipedia, the free encyclopedia.

- May 7th 2010, 09:08 AM #6

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