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

1 Answer
Aug 29, 2015

#x= (1+sqrt(1+4e))/2 = "(approx.)" 2.22287#

Explanation:

If
#color(white)("XXX")ln(x)+ln(x-1)=1#
then
#color(white)("XXX")e^(ln(x)+ln(x-1))= e^1#

#rarrcolor(white)("XXX")e^(ln(x))*e^(ln(x-1)) = e^1#

#rarrcolor(white)("XXX")x*(x-1) = e#

#rarrcolor(white)("XXX")x^2-x-e=0#

Using the quadratic formula
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

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

Since #(sqrt(1+4e) > 1)#
#color(white)("XXXXXXXXXXX")(1-sqrt(1+4e))/2# is negative
and
Since #ln(x)# is undefined for #x < 0#
#color(white)("XXX")x=(1-sqrt(1+4e))/2# is an extraneous solution
and only
#color(white)("XXX")x=(1+sqrt(1+4e))/2# is valid