Solving a simple logarithm

1. Jul 16, 2007

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!

2. Jul 16, 2007

ranger

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

3. Jul 16, 2007

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.

4. Jul 16, 2007

Kummer

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

5. Jul 16, 2007

HallsofIvy

Staff Emeritus
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.

6. Jul 16, 2007

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.

7. Jul 16, 2007

Dick

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

8. Jul 16, 2007

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 dont exponentiate yet, whats -ln(y)? Remeber a*log(x) = log(x^a) right? so whats (-1)*ln(y)?

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

9. Jul 16, 2007

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.

10. Jul 16, 2007

obsolete

I finally understand =)

Thanks all!

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