Answer
$x=\dfrac{1}{1+\log6}\approx0.562382$
Work Step by Step
$10^{1-x}=6^{x}$
Apply $\log$ to both sides of the equation:
$\log10^{1-x}=\log6^{x}$
The exponents $1-x$ and $x$ can be taken down to multiply in front of their respective logarithms:
$(1-x)\log10=x\log6$
Since $\log10=1$, the equation becomes:
$1-x=x\log6$
Solve for $x$:
$x\log6+x=1$
$x(\log6+1)=1$
$x=\dfrac{1}{1+\log6}\approx0.562382$