The problem with this equation lies with the denominator on the left hand side. If we recall our graph of ln(x) however, we know that ln(x) is always positive and not equal to 0. Now we can safely multiply it up. The equation now reads:3 = (ln(x))2 + 2ln(x)We can recognise this as a quadratic equation in ln(x), and factorise it as such:(ln(x)+3)(ln(x)-1) = 0from which we deduce the solutions exist where ln(x) is equal to 1 or -3, the latter of which does not exist for any real values. Hence we consider ln(x) = 1, which is achieved when x = e, our one and only real solution. (note that we can confirm this solution by substituting x = e into the original equation).