Question

Help understanding exponent laws:
I have an equation ln(x-y)=lnx+c. When I simplify a bit, I get 1-y=x+e^c, but the professor writes it as 1-y=cx. How is that possible?

Answer 1

sorry typo.... ln(1-y)**

Answer 2

ln(1-y)=lnx+c ..... somehow simplifies to 1-y=cx. I am getting 1-y=x+e^c

Answer 3

Well two things here, you can rewrite \[c= \ln (e^c)\] and it will combine like this:
\[\ln(1-y)=\ln x+\ln e^c = \ln (e^c * x)\]
\[1-y = e^c x\]

of course since \(e^c\) is arbitrary, just making it your new constant doesn't matter.

Answer 4

If log rules make you uneasy, I suggest just doing it this way instead:

\[\ln( 1-y) = \ln x + c\]
raise these as exponents:

\[e^{\ln(1-y)} = e^{\ln x + c}\]
then you can separate out the exponents with exponent rules instead of log rules:

\[e^{\ln(1-y)} = e^{\ln x + c} = e^{\ln x }e^c\]
And then you get the same answer