jmc

To find $x$ such that $3^x=\frac{1}{\sqrt{3}}$, notice that multiplying the numerator and denominator of $\frac{1}{\sqrt{3}}$ by $\sqrt{3}$ gives us $\frac{\sqrt{3}}{3},$ and factoring $\frac{\sqrt{3}}{3}$ gives us $\sqrt{3}\cdot \frac{1}{3},$ which is equal to $3^{\frac{1}{2}}\cdot 3^{-1}.$ Looking back at our original equation, this means that $3^x=3^{\frac{1}{2}}\cdot 3^{-1}=3^{\frac{1}{2}+-1},$ and therefore $x=\frac{1}{2}+-1=-\frac{1}{2}.$ Since $3^{-\frac{1}{2}}=\frac{1}{\sqrt{3}},$ ${\log }_3\frac{1}{\sqrt{3}}=\boxed{-\frac{1}{2}}.$