Logarithms - Get the second solution to a logarithm equation 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?

Your solution is incorrect because :
$\ln(11x-10) +(\ln(11x-10))^2$ is not equal to $\ln(11x-10 \times {(11x-10)}^2)$
To solve 
$$\ln(11x-10) + (\ln(11x-10))^2 = 6$$
Let $t=\ln(11x-10)$
$$t^2 +t -6=0$$
this gives , $t=2$ and $t=-3$
when $t=2 \longrightarrow \ln(11x-10)=2 \longrightarrow x=\frac{e^2+10}{11}$
when $t=-3 \longrightarrow \ln(11x-10)=-3 \longrightarrow x=\frac{10e^3+1}{11e^3}$
