Hi guys! Do you know how to solve x(e^x)+1 = 0? Can I use log? Thanks!

Question

saifoo.khan · Answer

@ChrisS

jhonyy9 · Answer

yes you need to us ln

jhonyy9 · Answer

- so because there is e^x

anonymous · Answer

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 · Answer

do you know the property of logarithm ?

than loga(a) =1   - yes ?

jhonyy9 · Answer

- and you know that ln is logarithm base e   - yes ?

anonymous · Answer

lhs is always +....

anonymous · Answer

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 · Answer

"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 · Answer

Ahhh. After graphing, how did you get x? :)

beginnersmind · Answer

Did _you_ try to graph them? :) If there's a solution it's where the two graphs intersect.

beginnersmind · Answer

To clarify, I don't have a solution. But graphing those two functions gives a pretty good idea of what's going on.

anonymous · Answer

Look, just differentiate xe^x and go from there.....