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    How do you solve:
    ln(x2+5x+6) - ln(x2+3x)=3
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    Use the rule:
    ln(a) - ln(b) ≡ ln(a/b)
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    Thanks, however the answer states that the answer is x=2/e3 - 1??
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    (Original post by computerwizz)
    How do you solve:
    ln(x2+5x+6) - ln(x2+3x)=3
    Applying log rules, you get ln[(x2+5x+6)/(x2+3x)]=3
    Factorise (x2+5x+6) and you get (x+2)(x+3).
    Factorise (x2+3x) and you get x(x+3)
    Therefore x2+5x+6)/(x2+3x)=(x+2)/x as common factors eliminate.

    ln[(x+2)/x]=3

    3=ln(e^3)

    This gives (x+2)/x = e^3.
    EDIT: I accidentally multiplied this out. It should be 1+(2/x)=e^3. You can then rearrange.
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    (Original post by JemmaSimmons)
    Applying log rules, you get ln[(x2+5x+6)/(x2+3x)]=3
    Factorise (x2+5x+6) and you get (x+2)(x+3).
    Factorise (x2+3x) and you get x(x+3)
    Therefore x2+5x+6)/(x2+3x)=(x+2)/x as common factors eliminate.

    ln[(x+2)/x]=3

    3=ln(e^3)

    This gives (x+2)/x = e^3.
    x^2+2x=e^

    x^2+2x-e^3 =0.

    Use quadratic formula to solve.
    Thanks, however the answer states that it
    is x=2/e3 - 1??
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    (Original post by jbiard)
    ln(x^2+5x+6) - ln(x^2+3x)=3
    ln( (x^2+5x+6)/(x^2+3x) ) =3
    (x^2+5x+6)/(x^2+3x)=e^3

    (x^2+5x+6)= e^3(x^2+3x)

    (1-e^3)x^2 + (5-3e^3)x +6 = 0

    X= 0.104791393 ,
    Don't do this ever again. Never post full answers on TSR. Delete it before you receive a notice.
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    (Original post by JemmaSimmons)
    Is that 2 over (e^3-1)?
    Yes!
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    ln(x^2+5x+6) - ln(x^2+3x)=3
    ln(X+3)(x+2)/x(X+3)=3
    (X+2)/x=e^3
    x+2=xe^3
    2=xe^3-x
    factorise and rearrange
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    (Original post by Rohit_Rocks10)
    Don't do this ever again. Never post full answers on TSR. Delete it before you receive a notice.
    woah why??
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    (Original post by jbiard)
    woah why??
    Much better to hint the person to the right answer rather than give them the full solution.
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    (Original post by TasnimCh)
    ln(x^2+5x+6) - ln(x^2+3x)=3
    ln(X+3)(x+2)/x(X+3)=3
    (X+2)/x=e^3
    x+2=xe^3
    2=xe^3-x
    factorise and rearrange

    Thanks!
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    (Original post by RDKGames)
    Much better to hint the person to the right answer rather than give them the full solution.
    cool, but why would i get a notice?!
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    (Original post by jbiard)
    cool, but why would i get a notice?!
    Because it's part of the maths forum's rules not to give out a full solution unless OP doesn't understand after a suitable number of hints.
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    (Original post by computerwizz)
    Thanks!
    No worries dude
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    (Original post by RDKGames)
    Because it's part of the maths forum's rules not to give out a full solution unless OP doesn't understand after a suitable number of hints.
    yeah i just read the maths forum rules, didnt realise you werent aloud to give full worked solutions before, sorry!
 
 
 
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