# Log

• Jan 6th 2013, 03:24 AM
srirahulan
Log
$\log_2 3+\log_3 4+\log_4 5+\log_5 6+\log_6 7+\log_7 8$ How to find the value for this?

Find the range for x?

$3x+4
• Jan 6th 2013, 06:14 AM
skeeter
Re: Log
what have you done with either of these questions?
• Jan 6th 2013, 04:22 PM
srirahulan
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>>
• Jan 6th 2013, 04:37 PM
skeeter
Re: Log
Quote:

Originally Posted by srirahulan
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>>

• Jan 6th 2013, 05:03 PM
srirahulan
Re: Log
First One,

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

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

And then What Can i Do?

Second One,
$3x+4

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

Case 2,
$x^2-6x<9-2x$
$x^2-4x-9<0$
$x=\frac{4 \pm \sqrt{16+36}}{2}$
then i get answers in roots so how can i find the range of x?
• Jan 6th 2013, 05:04 PM
Plato
Re: Log
Quote:

Originally Posted by srirahulan
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 $\sum\limits_{k = 2}^7 {\frac{{\ln (k + 1)}}{{\ln (k)}}}$ .

Using a CAS I get 7.276.
• Jan 6th 2013, 05:29 PM
srirahulan
Re: Log
I can't understand you solution or the first one..
• Jan 6th 2013, 05:36 PM
skeeter
Re: Log
Quote:

Originally Posted by srirahulan
Second One,
$3x+4

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

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

$3x+4

case (1) ...

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

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

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

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

case (2) ...

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

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

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

$-1.60555... < x < 5.60555...$

intersect the two solution sets
• Jan 6th 2013, 09:39 PM
srirahulan
Re: Log