<p>This isn't exactly calculus, but it was one of the past problems on the Calculus AB exam.</p>
<p>How do you find the inverse of ln [x/(x-1)]?</p>
<p>y = ln(x/x-1)</p>
<p>Solve for x in terms of y.</p>
<p>e^y = x/x-1
e^y<em>x - e^y = x
(e^y-1)</em>x - e^y = 0
x = e^y/e^y-1</p>