Math Help - ln(1/a) = -ln(a)

1. ln(1/a) = -ln(a)

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

by using fundermental theory of calculus?

2. Originally Posted by ShaXar
how can we prove ln(1/a) = -ln(a)

by using fundermental theory of calculus?
Look at the definition of $\ln$ (presumably you have a definition using calculus) then derive the laws of logarithms and apply them

CB

3. Originally Posted by ShaXar
how can we prove ln(1/a) = -ln(a)

by using fundermental theory of calculus?

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

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

Tonio

4. Originally Posted by ShaXar
how can we prove ln(1/a) = -ln(a)

by using fundermental theory of calculus?
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 ???

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

6. Originally Posted by Khalfan
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 ???
No he does not, what he needs to do is read tonio's post (not that I would be so explicit in explaining it myself obviously).

CB