Solving Logarithms: Finding the Solution for ln(x) + ln(y) = 0

[obsolete]

Hi everyone.

I know this question is quite simple but I can't wrap my head around it at the moment..

Solve for y: ln(x) + ln(y) = 0

I've tried differentiating both terms and then arranging for y but I get y = -x. The answer is meant to be y = 1/x.

Thanks all!

 

[ranger]

How about the property of:
ln(x) + ln(y) = ln(xy)

 

[Gib Z]

Taking ln x to the other side and exponentiating gives:

$$y = e^{-\ln x}$$

I think the problem you are having is that you are thinking "$e^{\ln x}= x$, so this one should be -x". But remember the the negative sign is in the exponent.

If that is still not clear enough perhaps this will help : $a \log x = \log (x^a)$ for any base.

 

[Kummer]

I've tried differentiating both terms and then arranging for y but I get y = -x. The answer is meant to be y = 1/x.

You cannot differentiate both sides, because those are numbers not functions. The derivatives will yield zero and nothing important.

 

[HallsofIvy]

How about the property of:
ln(x) + ln(y) = ln(xy)

Gosh, do I feel dumb! I did that by writing it as ln x= -ln y.

Yes, ln x+ ln y= ln xy= 0 is easier.

 

[obsolete]

Thanks for the replies everyone.

I had already considered ln(x) + ln(y) = ln(xy) but I don't know how to approach the problem using this law. Any directions?

Thanks again.

 

[Dick]

Thanks for the replies everyone.

I had already considered ln(x) + ln(y) = ln(xy) but I don't know how to approach the problem using this law. Any directions?

Thanks again.

You've got ln(xy)=0. What then is the value of xy?

 

[K.J.Healey]

Forget the ln(xy), and use your ln(x)+ln(y)=0
do what you did next, your gut instinct was right.
ln(x)=-ln(y)
now don't exponentiate yet, what's -ln(y)? Remeber a*log(x) = log(x^a) right? so what's (-1)*ln(y)?

After you understand that, exponentiate, and solve :) e^log(a)=a

 

[K.J.Healey]

Just to add, the ln(xy) way works too. Its just that I don't want you to think that the way you started was wrong.

 

[obsolete]

I finally understand =)

Thanks all!