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Answer 1 · 2017-11-14T01:37:19

#x# is either #1.729# or #-2.729#.

Use the rule of natural logarithms: #ln(a)+ln(b)=ln(a*b)#.

#ln(x-1) + ln(x+2) = 1#
#ln((x-1)*(x+2))=1#
#ln(x^2+x-2)=1#

We all know that #ln_x x=1#. Since the base here is #e#, and the equation is #ln_e (x^2+x-2)=1#, #e=x^2+x-2#.

#x^2+x-2=e# (Take #e# as #2.718#)
#x^2+x-2=2.718#
#x^2+x-2-2.718=0#
#x^2+x-4.718=0#

Here, #a=1, b=1, c=-4.718#

Use the quadratic formula: #(-b+-sqrt(b^2-4ac))/(2a)#

Solve:

#(-1+-sqrt(1^2-4*1*-4.718))/(2)#

#(-1+-sqrt(1-4*-4.718))/(2)#

#(-1+-sqrt(1+18.872))/(2)#

#(-1+sqrt(19.872))/(2),(-1-sqrt(19.872))/(2)#

#(-1+4.458)/(2),(-1-4.458)/(2)#

#(3.458)/(2),(-5.458)/(2)#

#1.729,-2.729( 3 s.f.)#

So, #x# is either #1.729# or #-2.729#.