# ln

• Oct 22nd 2007, 10:55 PM
taurus
ln
Hi

I got the following question:

make y the subject of the following:
ln(x) + ln(y) = 0

This is what I have done so far, could someone help please?:

ln(x) + ln(y) = 0
ln(y) = -ln(x)
e^(ln(y) = e^(-ln(x))
y = e^(-ln(x))

Does that seem right?
• Oct 22nd 2007, 11:03 PM
CaptainBlack
Quote:

Originally Posted by taurus
Hi

I got the following question:

make y the subject of the following:
ln(x) + ln(y) = 0

This is what I have done so far, could someone help please?:

ln(x) + ln(y) = 0
ln(y) = -ln(x)

So:

$\displaystyle \exp(\ln(y))=\exp(-\ln(x)) = [\exp(\ln(x))]^{-1}$

so:

$\displaystyle y=1/x$

RonL
• Oct 22nd 2007, 11:09 PM
taurus
so basically:

e^(-ln(x)) = 1/x

?
• Oct 22nd 2007, 11:13 PM
Jhevon
why was writing both sides in base e necessary? could we have just droped the logs when appropriate?

$\displaystyle \ln y + \ln x = 0$

$\displaystyle \Rightarrow \ln y = - \ln x$

$\displaystyle \Rightarrow \ln y = \ln \left( x^{-1} \right)$

$\displaystyle \Rightarrow y = x^{-1}$

OR

$\displaystyle \ln y + \ln x = 0$

$\displaystyle \Rightarrow \ln xy = 0$

$\displaystyle \Rightarrow xy = e^0 = 1$

$\displaystyle \Rightarrow y = \frac 1x$

Oh well, I guess it was just a matter of taste
• Oct 22nd 2007, 11:14 PM
Jhevon
Quote:

Originally Posted by taurus
so basically:

e^(-ln(x)) = 1/x

?

yes

$\displaystyle e^{- \ln x} = e^{\ln \left( x^{-1} \right)} = \frac 1x$

since $\displaystyle a^{\log_a b} = b$