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anonymous
 3 years ago
solve ln x= 1 ln (x+2)
anonymous
 3 years ago
solve ln x= 1 ln (x+2)

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anonymous
 3 years 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
 3 years 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.\[\large x^2+bx\]

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

blurbendy
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ahhh sorry website froze :( erased all my stuff...

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

anonymous
 3 years 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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