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Thread: Log

  1. #1
    Member srirahulan's Avatar
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    Talking Log

    $\displaystyle \log_2 3+\log_3 4+\log_4 5+\log_5 6+\log_6 7+\log_7 8$ How to find the value for this?

    Find the range for x?

    $\displaystyle 3x+4<x^2-6x<9-2x$
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  2. #2
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    Re: Log

    what have you done with either of these questions?
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  3. #3
    Member srirahulan's Avatar
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    Re: Log

    first one i can't done more but in second one i solve finally i get answers in root then i can't do more>>
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  4. #4
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    Re: Log

    Quote Originally Posted by srirahulan View Post
    first one i can't done more but in second one i solve finally i get answers in root then i can't do more>>
    show your complete solution ...
    Thanks from topsquark
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  5. #5
    Member srirahulan's Avatar
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    Arrow Re: Log

    First One,

    $\displaystyle \log_2 3+\log_3 4+\log_4 5+\log_5 6+\log_6 7+\log_7 8$

    $\displaystyle \log_2 3+\frac{2}{\log_2 3}+\frac{1}{\log_5 4}+\log_5 6+\frac{1}{\log_7 6}+\log_7 8$

    And then What Can i Do?


    Second One,
    $\displaystyle 3x+4<x^2-6x<9-2x$

    Case 1,
    $\displaystyle 3x+4<x^2 -6$
    $\displaystyle 0<x^2-3x-10$
    $\displaystyle 0<(x-5)(x+2)$
    In this case i get $\displaystyle x>5 $Or$\displaystyle x<-2$

    Case 2,
    $\displaystyle x^2-6x<9-2x$
    $\displaystyle x^2-4x-9<0$
    $\displaystyle x=\frac{4 \pm \sqrt{16+36}}{2}$
    then i get answers in roots so how can i find the range of x?
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  6. #6
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    Re: Log

    Quote Originally Posted by srirahulan View Post
    first one i can't done more but in second one i solve finally i get answers in root then i can't do more>>
    I see no simple way to do the first one.

    It is equivalent to $\displaystyle \sum\limits_{k = 2}^7 {\frac{{\ln (k + 1)}}{{\ln (k)}}} $ .

    Using a CAS I get 7.276.
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  7. #7
    Member srirahulan's Avatar
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    Re: Log

    I can't understand you solution or the first one..
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  8. #8
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    Re: Log

    Quote Originally Posted by srirahulan View Post
    Second One,
    $\displaystyle 3x+4<x^2-6x<9-2x$

    Case 1,
    $\displaystyle 3x+4<x^2 -6\color{red}x?$
    $\displaystyle 0<x^2-3x-10$
    $\displaystyle 0<(x-5)(x+2)$
    In this case i get $\displaystyle x>5 $Or$\displaystyle x<-2$

    Case 2,
    $\displaystyle x^2-6x<9-2x$
    $\displaystyle x^2-4x-9<0$
    $\displaystyle x=\frac{4 \pm \sqrt{16+36}}{2}$
    then i get answers in roots so how can i find the range of x?
    $\displaystyle 3x+4<x^2-6x<9-2x$

    case (1) ...

    $\displaystyle 3x+4 < x^2-6x$

    $\displaystyle 0 < x^2-9x-4$

    $\displaystyle x < \frac{9-\sqrt{97}}{2} \approx -0.4244...$

    $\displaystyle x > \frac{9+\sqrt{97}}{2} \approx 9.4244...$

    case (2) ...

    $\displaystyle x^2-6x < 9-2x$

    $\displaystyle x^2-4x-9 < 0$

    $\displaystyle 2-\sqrt{13} < x < 2+\sqrt{13}$

    $\displaystyle -1.60555... < x < 5.60555...$

    intersect the two solution sets
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  9. #9
    Member srirahulan's Avatar
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    Re: Log

    Thank you for the reply>><<
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