I think I solved this correctly, but could use an extra set of eyes.

$\displaystyle ln(3 \sqrt{x}) = \sqrt{ln x}$.

Working on the left:

$\displaystyle ln (3 \sqrt{x}) = ln 3 + ln \sqrt{x} = ln 3 + \frac{1}{2}ln{x}$

so now I have:

$\displaystyle ln 3 + \frac{1}{2}ln{x} = \sqrt{ln x}$

I made the substitution $\displaystyle u = \sqrt{ln x}$, so $\displaystyle u^2 = ln x$. Performing this substitution gives:

$\displaystyle ln 3 + \frac{1}{2}u^2 = u$.

Moving everything to one side:

$\displaystyle \frac{1}{2}u^2 - u + ln 3 = 0$.

A quadratic in u, so using the quadratic formula with a = .5, b = -1, c = ln 3, my discriminant is:

$\displaystyle 1 - 2 ln 3$.

My discriminant is negative, so I have no real solution for u, and thus no real solution for the whole thing.

Look okay? The way the problem was phrased in the text really suggests there's an answer, but...