How do you solve #Ln(x+1)-ln(x-2)=lnx^2#?

1 Answer
Apr 10, 2018

Answer:

approximately:
#x=2.5468#

Explanation:

#ln^[(x+1)/(x-2)]# = #ln^(x^2)#
we can cancel out the (Ln) parts and the exponents would be left out;
#(x+1)/(x-2)# = #x^2#
#x+1# = #x^2.(x-2)#
#x+1# = #x^3-2x^2#
#x^3#-#2x^2#-#x#-#1#=#0#
#x=2.5468#