Simplify Equations: x - 3y + (1 + ln3) = 0 | Step-by-Step Solution

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In summary, the conversation was about a problem involving the equation y - 1/3 = 1/3 (x - ln(1/3)), with a solution attempt and a discrepancy with the textbook's answer. The expert summarizer explains that the two answers are equivalent, with the difference being the use of the logarithmic rule ln(1/x) = -ln(x).
  • #1
nesan
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Homework Statement



y - 1 / 3 = 1 / 3 (x - ln(1/3))

The Attempt at a Solution



y - 1 / 3 = 1 / 3 (x - ln(1/3))

0 = 1 /3 x - 1/3 ln(1 / 3) + 1 / 3 - y

0 = x - ln(1 / 3) + 1 - 3y

I don't see any problem with it.

Text book says it's wrong. >_<

The answer is = x - 3y + (1 + ln3) = 0
 
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  • #2
You have pretty much the same thing, except that they have used

[tex]\text{ln}( 1/x ) = \text{ln}( x^{-1} ) = -\text{ln}(x)[/tex]

to convert [itex]\text{ln}(1/3)[/itex] to [itex]-\text{ln}(3)[/itex].
 
  • #3
nesan said:

Homework Statement



y - 1 / 3 = 1 / 3 (x - ln(1/3))

The Attempt at a Solution



y - 1 / 3 = 1 / 3 (x - ln(1/3))

0 = 1 /3 x - 1/3 ln(1 / 3) + 1 / 3 - y

0 = x - ln(1 / 3) + 1 - 3y

I don't see any problem with it.

Text book says it's wrong. >_<

The answer is = x - 3y + (1 + ln3) = 0
Your result and the book answer are equivalent.

ln(1/3) = ln(3-1) = (-1) ln(3)
 
  • #4
Steely Dan said:
You have pretty much the same thing, except that they have used

[tex]\text{ln}( 1/x ) = \text{ln}( x^{-1} ) = -\text{ln}(x)[/tex]

to convert [itex]\text{ln}(1/3)[/itex] to [itex]-\text{ln}(3)[/itex].

Thank you very much. :]

SammyS said:
Your result and the book answer are equivalent.

ln(1/3) = ln(3-1) = (-1) ln(3)
Thank you very much. :]
 

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