Re-writing a logarithm to a power

Given:

$$(4\ln x)^2$$

Is this simplified to $8\ln x$, (multiplying the expression by 2),
$32\ln x$, (square $4$ ($16$), then $\ln x$ ($2\ln x$) and combine again),
or something else?

Just to be sure, the equation I was solving for was $e^\sqrt t$ = $x^4$ for $t$.
My thought process is:

$\ln(x^4)$ = $\sqrt t$
$4\ln x$ = $\sqrt t$
$(4\ln x)^2$ = $t$

-
If the logarithm itself is squared, that's as simple as it can get. If what you had instead was $\ln x^2$, then something can be done... – J. M. Sep 21 '11 at 1:40
So is that already in simplest form? (Also, would you mind checking the edit I added to the bottom, just to ensure that I'm doing this right. I'm attempting to remember how to use logorithms at this point, so I'm still a beginner). – Mike Gates Sep 21 '11 at 1:42
The expression $(\ln x)^2$ is the same as $\ln x \cdot \ln x$. The expression $2\ln x$ is the same as $\ln x + \ln x$. Multiplying two numbers and adding two numbers should, in general, give different results. – Austin Mohr Sep 21 '11 at 1:48
Well done. One might want to rewrite the answer as $16(\ln x)^2$. – André Nicolas Sep 21 '11 at 5:26
If you really want to get rid of the exponent two you make things more complicated: $(\ln x)^2 = (\ln x)(\ln x) = \ln(x^{\ln x})$... – Dirk Sep 21 '11 at 6:19

It might be helpful to “take out” the logarithm for a second so you don’t get hung up on log properties. Let’s take $a = 4$ and $b = \ln x$; then
$$(4 \ln x)^2 = (ab)^2 = a^2 \cdot b^2 = 4^2 \cdot (\ln x)^2$$
It’s pretty clear now what the coefficient is going to be, and as J.M. pointed out above, there’s nothing more to be done with a term like $(\ln x)^2$.
As for the underlying question, you can substitute $\sqrt{t}$ into your expression easily enough; then just apply your exponential/logarithmic rules and check that the two sides are equal.