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

3 Answers
Oct 30, 2017

#(3+sqrt(13))/2#

Explanation:

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

If:

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

Then:

#(x+1)/(x-2)=x =x^2-3x-1#

Solving: #x^2-3x-1#

#x = (3+sqrt(13))/2 and x= (3-sqrt(13))/2#

# (3-sqrt(13))/2# not valid as gives negative value in #ln(x-2)#

So solution is:

#(3+sqrt(13))/2#

Oct 30, 2017

#x = 3/2 +sqrt(13)/2#

Explanation:

Taking the exponent of both sides, we get:

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

Multiplying both sides by #(x-2)# we get:

#x+1 = x^2-2x#

Subtract #x+1# from both sides to get:

#0 = x^2-3x-1#

#color(white)(0) = (x-3/2)^2-9/4-1#

#color(white)(0) = (x-3/2)^2-(sqrt(13)/2)^2#

#color(white)(0) = (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)#

So:

#x = 3/2 +-sqrt(13)/2#

One of these values #3/2+sqrt(13)/2# is positive and a valid solution of the original equation.

The other value #3/2-sqrt(13)/2# is negative so not a solution if #ln# is the real logarithm. In fact it is not even a solution if we consider the complex logarithm, for which:

#ln(3/2-sqrt(13)/2+1) - ln(3/2-sqrt(13)/2-2)#

#= ln(5/2-sqrt(13)/2) - ln(1/2+sqrt(13)/2)-ipi#

#!= ln(sqrt(13/2)-3/2)+ipi = ln(3/2-sqrt(13)/2)#

Oct 30, 2017

See below.

Explanation:

In the quest for real solutions, supposing #x ne {0,2}#

#log_e abs(x+1)-log_e abs(x-2) = log_e absx# or

#log_e(abs(x+1)/(abs(x-2)absx))=0# or

#abs(x+1)/(abs(x-2)absx)=1# whose solutions are included in the solutions for

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

#(x+1-(x-2)x)(x+1+(x-2)x)=0# or

#(1+3x-x^2)(1-x+x^2)=0#

and the real solutions are from

#1+3x-x^2=0# with the only feasible solution

#x = 1/2 (3 + sqrt[13])#