Thread: ln

**taurus** · October 22nd 2007, 10:55 PM

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?

**CaptainBlack** · October 22nd 2007, 11:03 PM

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:

$<br /> \exp(\ln(y))=\exp(-\ln(x)) = [\exp(\ln(x))]^{-1}<br />$

so:

$<br /> y=1/x<br />$

RonL

**taurus** · October 22nd 2007, 11:09 PM

so basically:

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

?

**Jhevon** · October 22nd 2007, 11:13 PM

why was writing both sides in base e necessary? could we have just droped the logs when appropriate?

$\ln y + \ln x = 0$

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

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

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

OR

$\ln y + \ln x = 0$

$\Rightarrow \ln xy = 0$

$\Rightarrow xy = e^0 = 1$

$\Rightarrow y = \frac 1x$

Oh well, I guess it was just a matter of taste

**Jhevon** · October 22nd 2007, 11:14 PM

Originally Posted by taurus

so basically:

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

?

yes

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

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

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