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