Question

How do you factor 2e^x - 2xe^x - 1 = 0?
Can you take ln of individual terms?

Answer 1

I dont think it can be factored

Answer 2

@ganeshie8 can u help?

Answer 3

doesnt look factorable... and solving analytically looks painful hmm

Answer 4

but you can solve for x here

Answer 5

uh...\[2e^x(1-x)=1\]

Answer 6

@mark_o. hw did u get that.. care to explain ?

Answer 7

I have no idea how you got that. I get stuck at \[e^x(x-1)=-\frac{1}{2}\]

Answer 8

even me.. i i take ln both sides, x-1 is going inside ln.. . i get stuck

Answer 9

yup
2e^x - 2xe^x - 1 = 0 factoring
2 (1-x) -1 =0
e^x (1-x)=1/2 take ln on both sides
ln [e^x (1-x)]= ln1/2
lne^x + ln(1-x)=ln1/2
x+ ln(1-x)=ln1/2

Answer 10

oops sorry i thought that was lne instead of lne^x

Answer 11

ohhkk :)

Answer 12

ok lets have fun helping some more here guys lol....