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

Question

1120 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-22T15:55:41

#x=1.729#

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

#hArrln(x-1)(x-2)=1# or

#(x-1)(x+2)=e# or

#x^2+x-2-e=0# and using quadratic fomula #(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-1+-sqrt(1^2-4xx1xx(-2-e)))/2#

= #(-1+-sqrt(1+8+4e))/2#

= #(-1+-sqrt(9+4e))/2#

= #(-1+-sqrt(9+4xx2.7183))/2#

= #(-1+-sqrt(9+10.8732))/2#

= #(-1+-sqrt(19.8732))/2#

= #(-1+-4.4579)/2#

i.e. #x=1.729# or #x=-2.729#

But #x# cannot have a negative value, hence #x=-2.729#