How do you solve #abs(x^2-2x-3) = sqrt (2- lnx)#?

The best way to solve this equation is using an iterative method, after knowing the initial trials. The initial iterations can be found after plotting the curves

#y_1 = abs(x^2-2x-3)# and
#y_2 = sqrt (2- lnx)#

this plot will cover the interval #0 < x < e^2# because in this interval we have feasible solutions.

We now will perform a series of algebraic transformations into the original problem, in order to remove handling difficulties.

First after squaring both equation terms, we consider

# abs(x^2-2x-3)^2= 2- log_ex#

and now we call

#f(x) = (x^2-2x-3)^2-2+log_e x = 0#

Supposing we know an initial guess #x_0# we can proceed using the Taylor series first approximation

#f(x_{k+1}) = f(x_k) + f'(x_k)(x_{k+1}-x_k) + O^2(x_k)#

but we need that #f(x_{k+1})= 0# then, supposing #O^2(x_k)# small enough, we obtain

#x_{k+1} = x_k - f(x_k)/(f'(x_k))#

From the plot we propose first #x_0=2#
and obtain successively
#x_1 = 2.66897#
#x_2 =2.72804#
#x_3 =2.73279#
#x_4 =2.73282#

and also #x_0 = 4#
obtaining successively
#x_1 = 3.59525#
#x_2 =3.35944#
#x_3 =3.24893#
#x_4 =3.21869#
#x_5 =3.21632#
#x_6 =3.21631#

This way we obtained the equation solution within 6 sd!

Attached a plot with #y_1# in red and #y_2# in blue, showing the two solutions.