Why is there no real solution: Ln4-Lnx=Ln4/(Lnx)

No, this isn't right at all. The left side is fine, but
[tex]\frac{\log a}{\log b} \ne \log{\left( \frac{a}{b} \right)}[/tex]
... which is what you did on the right side.

 

I don't believe that Jurassic meant to say that that was the entire equation- he was just simplifying the left side: ln(x)- ln(4)= ln(x/4)= ln(4)/ln(x) is the entire equation.

But, in fact, he should left it the way it was: ln(x)- ln(4)= ln(4)/ln(x). Now let y= ln(x) and the equation is y- ln(4)= ln(4)/y. Multiply through by y to get the quadratic equation [itex]y^2- ln(4)y- ln(4)= 0[/itex]. Solve that for y= ln(x), using the quadratic formula or by completing the square. Does that quadratic have positive real roots? Remember that the logarithm of any real number must be positive.