anonymous
  • anonymous
ln 2 - ln (3x + 2) = 1 Find X
Linear Algebra
schrodinger
  • schrodinger
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mathmale
  • mathmale
An important rule of logarithms is this : ln a - ln b = ln (a/b). Given that, what is ln 2 - ln (3x+2)?
anonymous
  • anonymous
ln 2/ (3x + 2)
mathmale
  • mathmale
good. Now, \[\ln \frac{ 2 }{ 3x+2 }=1.\] how would you now get rid of the ln operator? We want to solve this equation for x.

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anonymous
  • anonymous
e^ln ( 2/ 3x + 2 ) = e^1
mathmale
  • mathmale
Hey, that's really cool. Can you now simplify this equation?
anonymous
  • anonymous
no
anonymous
  • anonymous
i forgot how to
mathmale
  • mathmale
Use the fact that e^(ln a) = a. If this is true (which it is), then what is \[e ^{\ln \frac{ 2 }{ 3x+2 }}?\]
mathmale
  • mathmale
The right side is simply e^1, or e.
anonymous
  • anonymous
ok, so its now (2/3x + 2) = 1
mathmale
  • mathmale
Other than that your 3x+1 should be within parentheses, yes! Mind re-writing that? Don't need parenthesis in front of the first 2.
anonymous
  • anonymous
2/ (3x + 2) = e
mathmale
  • mathmale
Yes, or \[\frac{ 3x+2 }{ 2 }=\frac{ 1 }{ e }\] One way of solving this would be to cross-multiply. We have to isolate x on the left side of the equation. I need to log off now, but think you're well on the way to solving this problem properly. See you again soon, I hope.
anonymous
  • anonymous
im still confused
myininaya
  • myininaya
are you confused on the cross multiplication part? he means if you have \[\frac{a}{b}=\frac{c}{d} \text{ you can do } ad=cb\]
myininaya
  • myininaya
and remember we are solving for x remember treat e like you would treat 2 or 3 or any other constant number
anonymous
  • anonymous
ok ^_^ thanks

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