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camilasanchez
 one year ago
Best ResponseYou've already chosen the best response.0I 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I guess we would have to complete the square on the X's. Take half of the b term, and square it.\[\large x^2+bx\]

camilasanchez
 one year ago
Best ResponseYou've already chosen the best response.0so x^2+2x+1 = e ?

camilasanchez
 one year ago
Best ResponseYou've already chosen the best response.0but i get\[x= \sqrt{e}1\]

camilasanchez
 one year ago
Best ResponseYou've already chosen the best response.0is that the answer?

blurbendy
 one year ago
Best ResponseYou've already chosen the best response.2log(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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh sorry website froze :( erased all my stuff...

camilasanchez
 one year ago
Best ResponseYou've already chosen the best response.0thanks everyone and that sucks! @zepdrix :/ but thanks either way !

nitz
 one year ago
Best ResponseYou've already chosen the best response.0ln(x)+ln(x+2)=1 ln(x(x+2))=ln(e) x(x+2)=e \[x ^{2}+2xe=0\] D=4+4e=4(1+e) =>sqrtD=2sqrt(1+e) x1=[2+2sqrt(1e)]/2=1+sqrt(1+e)>0 accepted x2=[22sqrt(1e)]/2=1sqrt(1+e)<0 denied
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