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Math Help - Logarithme Help

  1. #1
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    Logarithme Help

    Hello Guys i have a logarithme lesson, and our teacher gave us some exercices but he didn't explain all the lesson.

    compare without a calculater:
    Ln7 and ln2 + ln 3
    For me: 7>3+5 --> Ln7>ln5

    3+ln (radical)e and 4 ? well i don't know how to compare it.


    Resolve this:
    lnx =1 ; ln 2x=0 ; lnx=-1 ; lnx=3 ( what i have to do here)??

    ln(x-3)=ln(x+7)-ln(x+1)??
    ln(x^2-2x-3) = ln(x+7)

    the teacher told us to do them and he gave us this
    ln(ab) = ln (a) +ln(b) but i can't do them....
    Please help me i have many exercices like these and when somebody could help me i guess i can solve the others ones

    Many thanks in advance
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by iceman1 View Post
    Ln7 and ln2 + ln 3
    \ln 2+\ln 3=\ln 6<\ln 7

    3+ln (radical)e and 4
    3+\ln\sqrt{e}=3+\frac{1}{2}\ln e=3+\frac{1}{2}<4

    Resolve this:
    lnx =1 ; ln 2x=0 ; lnx=-1 ; lnx=3
    Here you can use this: \log_ab=c\Leftrightarrow b=a^c


    ln(x-3)=ln(x+7)-ln(x+1)
    \ln(x-3)=\ln(x+7)-\ln(x+1)\Leftrightarrow\ln(x-3)+\ln(x+1)=\ln(x+7)\Leftrightarrow
    \Leftrightarrow\ln(x-3)(x+1)=\ln(x+7)\Leftrightarrow(x-3)(x+1)=x+7
    Now you can continue.
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  3. #3
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    Oh Thanks , I'll try to solve them all Now
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  4. #4
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    i solved this:
    Ln(2+ radical 3)^30 + Ln(2+radical 3)^3 my result is 0
    (x+2)Ln(x+2) i got x = -1



    But i need Help with this Guys

    Ln(x^2 - 4)=Ln5 + 2Ln3
    Ln(Radical) 3-x + Ln(radical)x+1=Ln(radical)10-6x --> should i make them ^2?

    system:
    x+y=30
    Lnx +lny=3Ln6


    i did the resr but one question how to find the derivebality of this
    Ln=[1-x^2]

    p.s For the Last question [ ] = sorry about the ''z'' LOL it's Nothing

    Thanks Now i am starting to love math and my teacher told my mum that i am contribution at class
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