## Algebra 2 Common Core

$x=2$
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: $\log{(5-2\cdot2)}=0$ $\log{(5-4)}=0$ $\log{(1)}=0$ $0=0$