Logarithm Equality $$\sqrt{\log_x\left(\sqrt{3x}\right)} \cdot \log_3 x = -1$$
I am not entirely sure how to go about solving for $x$. I cannot square each side because the product isn't $≥ 0$, I can't think of any more approaches right now.

Since square roots are nonnegative, we know that $\log_3 x < 0$ so that $0 < x < 1$. Now let $k = \log 3$ and let $y = \log x$. Then by using log rules, our equation becomes:
\begin{align*}
\sqrt{\frac{\log \sqrt{3x}}{\log x}} \cdot  \frac{\log x}{\log 3} &= -1 \\
\sqrt{\frac{\frac{1}{2}\log 3x}{\log x}} \cdot  \frac{\log x}{\log 3} &= -1 \\
\sqrt{\frac{\frac{1}{2}(\log 3 + \log x)}{\log x}} &= -\frac{\log 3}{\log x} \\
\sqrt{\frac{\frac{1}{2}(k + y)}{y}} &= -\frac{k}{y} \\
\frac{\frac{1}{2}(k + y)}{y} &= \frac{k^2}{y^2} \\
ky + y^2 &= 2k^2 \\
y^2 + ky + \frac{1}{4}k^2 &= \frac{9}{4}k^2 \\
\left(y + \frac{1}{2}k\right)^2 &= \frac{9}{4}k^2 \\
y + \frac{1}{2}k &= \pm \frac{3}{2}k \\
y &= k,-2k \\
\end{align*}
Converting back, we find that either $\log x = \log 3$ or $\log x = -2\log 3 = \log (1/9)$ so that either $x = 3$ or $x = 1/9$. But then since $0 < x < 1$, we reject the first extraneous solution and conclude that $x = 1/9$.
