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anonymous

  • one year ago

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?

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  1. anonymous
    • one year ago
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    sorry typo.... ln(1-y)**

  2. anonymous
    • one year ago
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    ln(1-y)=lnx+c ..... somehow simplifies to 1-y=cx. I am getting 1-y=x+e^c

  3. Empty
    • one year ago
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    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.

  4. Empty
    • one year ago
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    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

  5. mathstudent55
    • one year ago
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    |dw:1442120586554:dw|

  6. anonymous
    • one year ago
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    Thank you everyone. I am a little shaky on the exponent laws so that was helpful.

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