## College Algebra (11th Edition)

$x=-\frac{2}{3}$
We rewrite the right side, so that the base is $\frac{1}{e}$ on both sides. $\frac{1}{e}^{-x}=[(\frac{1}{e})^2]^{x+1}$ By the law of exponents: $(a^x)^y=a^{xy}$ $((\frac{1}{e})^2)^{x+1}=(\frac{1}{e})^{2x+2}$ The rewritten equation: $\frac{1}{e}^{-x}=(\frac{1}{e})^{2x+2}$ Then, since the bases are the same, we set the exponents equal, because the two sides are equal only if the powers are equal too: if $b^m=b^n$ and $b \ne0$ , $b \ne1$ then $m=n$ $-x=2x+2 \\-x-2x=2$ $-3x=2 \\\frac{-3x}{3}=\frac{2}{-3}$ $x=-\frac{2}{3}$