Answer
$x=2$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition to our equation (with $b=10, y=0, x=5-2x$), we get:
$10^{0}=5-2x$
$1=5-2x$
$1-5=-2x$
$-4=-2x$
$x=\frac{-4}{-2}$
$x=2$
We confirm that the answer works:
$\log{(5-2\cdot2)}=0$
$\log{(5-4)}=0$
$\log{(1)}=0$
$0=0$