anonymous
  • anonymous
Hi guys! Do you know how to solve x(e^x)+1 = 0? Can I use log? Thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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saifoo.khan
  • saifoo.khan
@ChrisS
jhonyy9
  • jhonyy9
yes you need to us ln
jhonyy9
  • jhonyy9
- so because there is e^x

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anonymous
  • anonymous
So ln (x) + ln (e^x) =ln (-1). If we ln both sides. But would that be wrong, since -1 is not in the domain for the natural log function?
jhonyy9
  • jhonyy9
do you know the property of logarithm ? than loga(a) =1 - yes ?
jhonyy9
  • jhonyy9
- and you know that ln is logarithm base e - yes ?
anonymous
  • anonymous
lhs is always +....
anonymous
  • anonymous
Well, I moved the +1 to the RHS so that ln of a product would be left on the LHS and thus, I can separate them into products. ln (ab)= ln a + ln b, right? Or am I forgetting something?
beginnersmind
  • beginnersmind
"lhs is always +...." True, but how do you prove it? greencat, I subtracted 1 and divided by x. Got e^x=-1/x and graphed both.
anonymous
  • anonymous
Ahhh. After graphing, how did you get x? :)
beginnersmind
  • beginnersmind
Did _you_ try to graph them? :) If there's a solution it's where the two graphs intersect.
beginnersmind
  • beginnersmind
To clarify, I don't have a solution. But graphing those two functions gives a pretty good idea of what's going on.
anonymous
  • anonymous
Look, just differentiate xe^x and go from there.....

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