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

1 Answer
Nov 24, 2015

Answer:

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

Explanation:

Before beginning, note that we do not accept values less than #0# for the #ln# function, and thus as we are beginning with #ln(x)# in the equation, it must be that #x > 0#.

We will use the following properties of logarithms:
(1) #ln(ab) = ln(a) + ln(b)#
(2) #e^(ln(a)) = a#

as well as the quadratic formula:
#ax^2 +bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)#


By (1)
#ln(x) + ln(x+1) = ln(x(x+1))=ln(x^2 + x)#
#=> ln(x^2 + x) = 2#

Then, by (2)
#e^2 = e^(ln(x^2+x)) = x^2 + x#
#=> x^2 + x - e^2 = 0#

Now we apply the quadratic formula with #a=b=1# and #c=-e^2#

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

But as noted at the beginning, we must have #x > 0# and so we reject #(-1-sqrt(1+e^2))/2#

Thus we are left with the answer

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