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?

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- Oct 22nd 2007, 10:55 PMtaurusln
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 PMCaptainBlack
- Oct 22nd 2007, 11:09 PMtaurus
so basically:

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

? - Oct 22nd 2007, 11:13 PMJhevon
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 PMJhevon