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

Find the range for x?

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

Results 1 to 9 of 9

- Jan 6th 2013, 02:24 AM #1

- Jan 6th 2013, 05:14 AM #2

- Jan 6th 2013, 03:22 PM #3

- Jan 6th 2013, 03:37 PM #4

- Jan 6th 2013, 04:03 PM #5
## 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?

- Jan 6th 2013, 04:04 PM #6

- Jan 6th 2013, 04:29 PM #7

- Jan 6th 2013, 04:36 PM #8
## Re: Log

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

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

- Jan 6th 2013, 08:39 PM #9