Answer
$x=-\dfrac{\log3-\log2}{\log2}\approx-0.584963$
Work Step by Step
$2^{1-x}=3$
We can write the left side of the equation like this:
$(2)(2^{-x})=3$
Take the $2$ to divide the right side of the equation. It becomes:
$2^{-x}=\dfrac{3}{2}$
Apply $\log$ to both sides of the equation:
$\log2^{-x}=\log{\dfrac{3}{2}}$
Now, the exponent $-x$ can be taken down to multiply in front of the logarithm:
$-x\log2=\log{\dfrac{3}{2}}$
Solve for $x$:
$-x=\dfrac{\log{\dfrac{3}{2}}}{\log2}$
$x=-\dfrac{\log{\dfrac{3}{2}}}{\log2}$
We know that the $\log$ of a division can be expanded as a substraction, so we can write the numerator like this:
$x=-\dfrac{\log3-\log2}{\log2}\approx-0.584963$