# How do i solve this problem?

• September 17th 2008, 06:45 PM
Diggidy
How do i solve this problem?
http://webwork.math.ttu.edu/webwork2...d2243a3111.png

Do i do : lnx^2-ln(1-x) = lnx + ln2x - ln (1+x)?

if so... what is my next step?
• September 17th 2008, 06:53 PM
skeeter
$\ln\left(\frac{x^2}{1-x}\right) = \ln\left(\frac{2x^2}{1+x}\right)
$

$\frac{x^2}{1-x} = \frac{2x^2}{1+x}$
• September 17th 2008, 07:04 PM
javax
Quote:

Originally Posted by Diggidy
http://webwork.math.ttu.edu/webwork2...d2243a3111.png

Do i do : lnx^2-ln(1-x) = lnx + ln2x - ln (1+x)?

if so... what is my next step?

find x?
if so here we go...
$\ln{\left(\frac{x^2}{1-x}\right)} = \ln{\left(\frac{x 2x}{1+x}\right)}
$

$\Rightarrow$
$\frac{x^2}{1-x}=\frac{2x^2}{1+x}$

now cross multiply and you'll find x
• September 17th 2008, 07:17 PM
Diggidy
got it, thank you