2^x = 3^(x+1)

Take the natural log of both sides (any base will do, actually).

ln[2^x] = ln[3^(x+1)] <--- log[a^x] = x*log(a) for all bases.

x*ln(2) = (x+1)*ln(3)

x*[ln(2) - ln(3)] = ln(3)

x = ln(3)/[ln(2) - ln(3)] = -ln(3)/[ln(3) - ln(2)]

(The last step isn't really necessary, I just find it to be a bit tidier since x is negative.)

-Dan