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Math Help - LOG QUESTIONs please help

  1. #1
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    Question LOG QUESTIONs please help

    need some help with these questions PLEASE

    1.
    solve the equation

    0.5 ^(1+3x)=7

    giging you answer to 2 decimal places

    2.
    log (5x) - log(1/x+1)=2log(2x+2)

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  2. #2
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    Quote Originally Posted by pal4499 View Post
    need some help with these questions PLEASE

    1.
    solve the equation

    0.5 ^(1+3x)=7

    giging you answer to 2 decimal places

    2.
    log (5x) - log(1/x+1)=2log(2x+2)

    to #1:

    \left(\dfrac12\right)^{1+3x} = 7~\iff~\dfrac12 \left(\dfrac18\right)^x=7~\implies~ \left(\dfrac18\right)^x=14~\implies~8^x=\dfrac1{14  }~\implies~x=\log_8\left(\dfrac1{14}\right)~\impli  es~  x=\log_8\left(14^{-1}\right)~\implies~x=\dfrac{-\ln(14)}{\ln(8)}

    to #2:

    Unfortunately the writing isn't clear. I assume that you mean:

    \log(5x)-\log\left(\dfrac1{x+1}\right)=2\log(2x+2)~\implies  ~\log(5x(x+1))=\log((2x+2)^2)

    Solve for x:

    5x(x+1)=(2x+2)^2~\iff~5x(x+1)=4(x+1)^2~\implies~x=-1~\vee~x=4

    Keep in mind that the log-function is defined for positive values only and therefore x = -1 is not a solution of this equation!


    EDIT: Removed a mistake
    Last edited by earboth; November 4th 2008 at 11:14 AM.
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  3. #3
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    i may have written the question out wrong!!

    (1)
    0.5(1+3x)=7

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  4. #4
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    Quote Originally Posted by pal4499 View Post
    i may have written the question out wrong!!

    (1)
    0.5(1+3x)=7

    Remember that .... 0.5=\dfrac12
    So I've done your question in my previous post. I didn't give you the approximative number because I was pretty sure that you can calculate it yourself. For confirmation x\approx -1.27

    EDIT: I've corrected my previous post. Have a look!
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