ln(x+1) = ln(3x+1) - ln x

 ln(x+1) + ln(x) = ln(3x+1)

      ln[(x+1)x] = ln(3x+1)

        ln(x²+x) = ln(3x+1)

Use the rule:  if ln(A) = ln(B) then A = B

            x²+x = 3x+1

     x² - 2x - 1 = 0

Get 0 on the right by subtracting 3x and 1 
from both sides:

     x² - 2x - 1 = 0

Use the quadratic formula:
                  ______ 
            -b ± Öb²-4ac
        x = —————————————
                2a 

where a = 1; b = -2; c = -1

                      ______________
             -(-2) ± Ö(-2)²-4(1)(-1)
        x = ————————————————————————
                     2(1) 
                  ___ 
             2 ± Ö4+4
        x = ———————————
                 2

                  _ 
             2 ± Ö8
        x = ————————
                2 

                  ___ 
             2 ± Ö4·2
        x = ———————————
                 2 

                   _ 
             2 ± 2Ö2
        x = ——————————
                2 

                     _
             2     2Ö2
        x = ——— ± —————
             2      2
                 _
        x = 1 ± Ö2 
                      _
Using the +, x = 1 + Ö2, which
is one answer and equals about 2.141213562
                      _ 
Using the -, x = 1 - Ö2, which
is the other answer and equals about -.4142135624.

However since the original problem contains ln(x), 
and since logarithms can only be taken of positive 
numbers, the only solution is:
          _
 x = 1 + Ö2  

Edwin