- #1
snowJT
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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 that's not right.. let me know where I went wrong