Simple exponents

Simply convert the base:

${ \quad \quad 3 }^{ 1-x }=\frac { 1 }{ \sqrt { 3 } } \\ \quad \quad \\ \quad \quad { 3 }^{ 1-x }={ 3 }^{ -\frac { 1 }{ 2 } }\\ \\ \Leftrightarrow 1-x=-\frac { 1 }{ 2 } \\ \\ \quad \quad \quad \quad x=\boxed { \frac { 3 }{ 2 } }$

$3^{1-x} = \frac{1}{\sqrt{3}}$ 1. We square both sides to remove the square root. $3^{2-2x} = \frac{1}{3}$ 2. Then we multiply by $3$ to get rid of the fraction $3^{3-2x} = 1$ 3. After that we use logarithms to simplify the question $\log_{3}{3^{3-2x}} = \log_{3}{1}$ $(3-2x)\log_{3}{3} = \log_{3}{1}$ $(3-2x) \cdot 1 = 0$ $3-2x = 0$ 4. Lastly we figure it out like a normal algebra question $3 = 2x$ $x = \boxed{\frac{3}{2}}$

$or$

1. Turn the root into a power of $3$ $3^{1-x} = \frac{1}{3^{\frac{1}{2}}}$
2. Combine the fraction and power into one power of $3$ $3^{1-x} = 3^{-\frac{1}{2}}$
3. Since the bases are the same they can be gotten rid of $1-x = -\frac{1}{2}$
4. Then we find $x$ normally $-x = -\frac{1}{2} - 1$ $-x = -1\frac{1}{2}$ $x = 1\frac{1}{2} = \boxed{\frac{3}{2}}$

3^1-x=3^1-3/2=3^-1/2=1/3^1/2, Therefore x=3/2