<p>My brain is fried...</p>
<p>lnx = 2 + ln (1-x) </p>
<p>thanks in advance</p>
<p>lnx = 2 + ln(1-x)
lnx - ln(1-x) = 2
ln(x/1-x) = 2 assuming both x and 1-x are positive
e^2 = x/1-x
...solve for x</p>
<p>I think that works.</p>
<p>ln x = 2 + ln(1-x)
ln x - ln (1-x) = 2
ln (x/(1-x)) = 2
x/(1-x) = e^2
x = e^2 - xe^2
x + xe^2 = e^2
x(1+e^2) = e^2
x = e^2/(1+e^2)</p>
<p>omg i feel so smart i knew how to do this like 2 months ago. yay. normally people ask like calc questions that i cant even comprehend! logarithms yay!</p>
<p>This forum needs LATEX, badly...it would make writing math a lot easier and more intelligible.</p>
<p>Heck yes.</p>
<p>Petition!</p>