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camilasanchez

  • 3 years ago

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

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  1. camilasanchez
    • 3 years ago
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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

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

  3. camilasanchez
    • 3 years ago
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    so x^2+2x+1 = e ?

  4. camilasanchez
    • 3 years ago
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    but i get\[x= \sqrt{e}-1\]

  5. camilasanchez
    • 3 years ago
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    is that the answer?

  6. blurbendy
    • 3 years ago
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    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

  7. zepdrix
    • 3 years ago
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    Ahhh sorry website froze :( erased all my stuff...

  8. camilasanchez
    • 3 years ago
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    thanks everyone and that sucks! @zepdrix :/ but thanks either way !

  9. nitz
    • 3 years ago
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    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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