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Help me find y

  • Thread starter snowJT
  • Start date
118
0
1. Homework Statement

If [tex]ln|y| - ln|y-a| = b[/tex] find y

2. The attempt at a solution

[tex]ln\frac{y}{y-a} = b[/tex]

[tex]\frac{y}{y-a} = e^b[/tex]

[tex]y-e^by = -ae^b[/tex]

[tex]y (-e^b) = -ae^b[/tex]

[tex]-y = \frac{-ae^b}{e^b}[/tex]

[tex]y = a[/tex]

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

cristo

Staff Emeritus
Science Advisor
8,056
72
1. Homework Statement

If [tex]ln|y| - ln|y-a| = b[/tex] find y

2. The attempt at a solution

[tex]ln\frac{y}{y-a} = b[/tex]

[tex]\frac{y}{y-a} = e^b[/tex]

[tex]y-e^by = -ae^b[/tex]

[tex]y (-e^b) = -ae^b[/tex]
Here's your mistake: this line should read [tex]y(1-e^b)=-ae^b [/tex]
 
118
0
thanks.. so...

[tex]y-e^by = -ae^b[/tex]

[tex]y = \frac{-ae^b}{y-e^b}[/tex]

[tex]y^2 = \frac{-ae^b}{-e^b}[/tex]

[tex]y^2 = a[/tex]

[tex]y = \sqrt a[/tex]
 
1,073
1
thanks.. so...

[tex]y-e^by = -ae^b[/tex]

[tex]y = \frac{-ae^b}{y-e^b}[/tex]
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.

[tex]1 = \frac{-ae^b}{y-e^b}[/tex]
 
763
0
thanks.. so...

[tex]y-e^by = -ae^b[/tex]

[tex]y = \frac{-ae^b}{y-e^b}[/tex]

[tex]y^2 = \frac{-ae^b}{-e^b}[/tex]

[tex]y^2 = a[/tex]

[tex]y = \sqrt a[/tex]
snowJT what are you doing? why did you divid by [tex]y-e^b[/tex] ?

continue from what cristo said then divid the right hand side by [tex]1-e^b[/tex]

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)
 

cristo

Staff Emeritus
Science Advisor
8,056
72
snowJT what are you doing? why did you divid by [tex]y-e^b[/tex] ?

continue from what cristo said then divid the right hand side by [tex]1-e^b[/tex]


then multiply top and bottom of the fraction on the right hand side by minus
This is correct.
and do ln of both sides and use the rules of log to get y
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!
 
763
0
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
 
118
0
oh, lol.. no I don't know what I was thinking..
 

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