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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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
I guess we would have to complete the square on the X's. Take half of the b term, and square it.\[\large x^2+bx\]
so x^2+2x+1 = e ?

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but i get\[x= \sqrt{e}-1\]
is that the answer?
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
Ahhh sorry website froze :( erased all my stuff...
thanks everyone and that sucks! @zepdrix :/ but thanks either way !
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

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