Do i do : lnx^2-ln(1-x) = lnx + ln2x - ln (1+x)? if so... what is my next step?
Follow Math Help Forum on Facebook and Google+
$\displaystyle \ln\left(\frac{x^2}{1-x}\right) = \ln\left(\frac{2x^2}{1+x}\right) $ $\displaystyle \frac{x^2}{1-x} = \frac{2x^2}{1+x}$
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... $\displaystyle \ln{\left(\frac{x^2}{1-x}\right)} = \ln{\left(\frac{x 2x}{1+x}\right)} $ $\displaystyle \Rightarrow$ $\displaystyle \frac{x^2}{1-x}=\frac{2x^2}{1+x}$ now cross multiply and you'll find x
got it, thank you
View Tag Cloud