## Precalculus (6th Edition) Blitzer

$x=4$
RECALL: (i) If $a^M=a^N$, then $M=N.$ (ii) $(a^m)^n=a^{mn}.$ Note that $\dfrac{1}{27}=\dfrac{1}{3^3} = 3^{-3}$. Thus, the given equation can be written as $3^{1-x} = 3^{-3}.$ Use rule (i) above to obtain $1-x=-3 \\-x = -3-1 \\-x=-4 \\(-1)(-x)=-4(-1) \\x=4.$