# Thread: How do i solve this problem?

1. ## How do i solve this problem?

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

if so... what is my next step?

2. $\ln\left(\frac{x^2}{1-x}\right) = \ln\left(\frac{2x^2}{1+x}\right)
$

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

3. Originally Posted by Diggidy

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

4. got it, thank you