# Homework Help: Help me find y

1. Jan 7, 2007

### snowJT

1. The problem statement, all variables and given/known data

If $$ln|y| - ln|y-a| = b$$ find y

2. The attempt at a solution

$$ln\frac{y}{y-a} = b$$

$$\frac{y}{y-a} = e^b$$

$$y-e^by = -ae^b$$

$$y (-e^b) = -ae^b$$

$$-y = \frac{-ae^b}{e^b}$$

$$y = a$$

I know thats not right.. let me know where I went wrong

2. Jan 7, 2007

### cristo

Staff Emeritus
Here's your mistake: this line should read $$y(1-e^b)=-ae^b$$

3. Jan 7, 2007

### snowJT

thanks.. so...

$$y-e^by = -ae^b$$

$$y = \frac{-ae^b}{y-e^b}$$

$$y^2 = \frac{-ae^b}{-e^b}$$

$$y^2 = a$$

$$y = \sqrt a$$

4. Jan 7, 2007

### d_leet

This is wrong you divided both sides by (y-eb) but you didn't divide correctly on the left hand side, it should be 1.

$$1 = \frac{-ae^b}{y-e^b}$$

5. Jan 8, 2007

### sara_87

snowJT what are you doing? why did you divid by $$y-e^b$$ ?

continue from what cristo said then divid the right hand side by $$1-e^b$$

then multiply top and bottom of the fraction on the right hand side by minus

and do ln of both sides and use the rules of log to get y

(if my advice is wrong or there is a simpler method some one please tell me because i actually got two answers and posted the one that might be more correct)

6. Jan 8, 2007

### cristo

Staff Emeritus
This is correct.
I don't see why you'd take the log. We are seeking y, and after performing the operations listed above, we have y on the left hand side!

7. Jan 8, 2007

### sara_87

oh, i thought (to get y in a simpler form) you could expand out the ln then do e again, but that wouldn't work and i should know that! (i think i'm getting tired!)

i don't think i should give out any advice next time unless i'm awake and certain, i just feel that i owe people lots of help because i got so much help from this site lol

8. Jan 9, 2007

### snowJT

oh, lol.. no I don't know what I was thinking..