Question

solve ln x= 1- ln (x+2)

Answer 1

I get :
lnx+ ln (x+2)
ln(x)(x+2)
ln (x^2 + 2x)= 1

e^1 = x^2 +2x ? and i need help from there on

Answer 2

I guess we would have to complete the square on the X's.
Take half of the b term, and square it.\[\large x^2+bx\]

Answer 3

so x^2+2x+1 = e ?

Answer 4

but i get\[x= \sqrt{e}-1\]

Answer 5

is that the answer?

Answer 6

log(x) = 1 - log(x + 2)
log(x) - 1 + log(x+2) = 0
log(x(x+2)) = 1
x(x+2) = e
x^2 + 2x = e
x^2 + 2x + 1 = 1 + e
(x + 1)^2 = 1 + e

x+1 = Sqrt[1 + e] or x + 1 = - Sqrt[1 + e}
x = Sqrt[1 + e] - 1 or x + 1 = - Sqrt[1 + e]

x = Sqrt[1 + e] - 1 or x = -1 -Sqrt[1 + e]
Substitute back into the original equation, only 1 will be right

x = Sqrt[1 + e] - 1

Answer 7

Ahhh sorry website froze :( erased all my stuff...

Answer 8

thanks everyone and that sucks! @zepdrix :/ but thanks either way !

Answer 9

ln(x)+ln(x+2)=1
ln(x(x+2))=ln(e)
x(x+2)=e
\[x ^{2}+2x-e=0\]
D=4+4e=4(1+e) =>sqrtD=2sqrt(1+e)
x1=[-2+2sqrt(1-e)]/2=-1+sqrt(1+e)>0 accepted
x2=[-2-2sqrt(1-e)]/2=-1-sqrt(1+e)<0 denied