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.

$$\log_x(\sqrt{3x})  = \log_x(3x)^{\frac{1}{2}} = \frac{1}{2} \log_x(3x)=(*)$$
To change the base of the logarithm, so that you have at both logarithms the base $3$, use the formula:
$$\log_b a=\frac{\log_c a}{\log_c b}$$
$$\log_x {3x}=\frac{\log_3 {3x}}{\log_3 {x}}=\frac{\log_3 {3}+\log_3 {x}}{\log_3 {x}}=\frac{1+\log_3 {x}}{\log_3 {x}}$$
$$(*)=\frac{1}{2} \frac{1+\log_3 {x}}{\log_3 {x}}=\frac{1}{2} \left ( \frac{1}{\log_3{x}}+1 \right )$$
$$\sqrt{\log_x(\sqrt{3x})} \cdot \log_3{x}=-1 \Rightarrow \sqrt{\frac{1}{2} \left ( \frac{1}{\log_3{x}}+1 \right )} \cdot \log_3{x}=-1 \\ \Rightarrow \sqrt{\frac{\log_3^2{x}}{2} \left ( \frac{1}{\log_3{x}}+1 \right )} =-1 \Rightarrow \sqrt{\frac{1}{2} \left ( \log_3{x}+\log_3^2{x} \right )} =\sqrt{i} \Rightarrow \frac{1}{2} \left ( \log_3{x}+\log_3^2{x} \right ) =i $$
Continue by solving for $x$.
