Find the Cartesian equation of the locus of the point $P$ which is the intersection of the tangent lines at a point $t$ on the graphs of $y=e^x$ and $y=\ln(x)$
My attempt
Equation of tangent at point $(t,e^t)$ on $y=e^x$
$y=e^tx+e^t(1-t)$
Equation of tangent at point $(t,\ln(t))$ on $y=\ln(x)$
$y=\frac{1}{t}x+\ln(t)-1$
So the intersection of the two lines is when
$(e^t-\frac{1}{t})x=\ln(t)-1-e^t(1-t)$
So the x coordinate of P is
$x=\frac{t\ln(t)-t+te^t-t^2e^t}{te^t-1}$
I plug this into one of the tangent line equations and I get a really messy value for the y coordinate of P.
I suspect that there is no Cartesian equation of the locus and that it has to be expressed parametrically but I am not sure since it is a question in my textbook. Is there another way to tackle this problem? Some help would be greatly appreciated.