Big doubt on small logarithmic equation

Question

I was given the equation $\sqrt{\ln{(x)}}=\ln (\sqrt{x})$.

I found two ways to go about this, and both seem to me reasonable. Clearly I'm wrong, since they yield different results. First way,

$\sqrt{\ln{(x)}}=\ln (\sqrt{x})$

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

$=\ln(x)=\frac{1}{2}\ln^2 (x)$

Let $u=\ln(x)$ and solve the quadratic expression; the results for $x$ are $x_1=1$, $x_2=e^4$. According to Symbolab this is the right answer.

But what if, instead of taking the square root out of the argument on the RHS and only after squaring both sides, we do it the other way around? This is, first square both members of the equation, and then apply logarithmic property. Like this:

$\sqrt{\ln{(x)}}=\ln (\sqrt{x})$

$=\ln(x)=\ln^2 (\sqrt{x})$

=$\ln(x)=\frac{1}{2}\ln^2(x)$

Again, let $u=\ln(x)$. As you can see, the coefficient of $u^2$ is now $\frac{1}{2}$, when on the other equation it ended up being $\frac{1}{4}$, thus yielding different result. But I can't see why this second way of solving the equation is wrong. Can somebody clear this up for me? Thanks.

Answer 1

$$ (\ln^{2}\sqrt{x}) = \left( \ln \sqrt{x} \right) \left( \ln \sqrt{x} \right) = \left( \frac{1}{2} \ln x \right) \left( \frac{1}{2} \ln x \right) = \frac{1}{4}(\ln x)^{2} = \frac{1}{4} \ln^{2} x.$$

Answer 2

You just made a little mistake in the last step. $\ln (\sqrt x)=\frac{1}{2} \ln (x)$ and when you square this you have to square $\frac{1}{2}$ also.

Answer 3

$$\sqrt{\ln x} = \ln (\sqrt x)$$

I think it would have been easier to get rid of the "ln"s.

Let $x=e^{2t}$ where $t \ge 0$. Then the equation becomes

\begin{align} \sqrt{\ln x} &= \ln (\sqrt x) \\ \sqrt{\ln e^{2t}} &= \ln \sqrt{(e^{2t})} \\ \sqrt{2t} &= t \\ 2t &= t^2 \\ t &\in \{0, 2\} \\ x \in \{1, e^4\} \end{align}