Here's the original equation:
$\ln(11x-10) + \Big(\ln(11x-10)\Big)^2$ = 6
I've managed to obtain one solution: $x = \frac{e^2 + 10}{11}$
through those steps:
$\ln\Big((11x-10)(11x-10)^2\Big) = 6$
$\ln\Big((11x-10)^3\Big) = 6$
$(11x-10)^3 = e^6$
$11x-10 = e^2$
$11x = e^2+10$
$x = \dfrac{e^2+10}{11}$
The textbook shows a second solution: $x = \dfrac{10e^3 + 1}{11e^3}$, but how do you get to that result?