## Intermediate Algebra (6th Edition)

$x=-\frac{1}{3}$
We are given that $27^{x+1}=9$. Both of these numbers are powers of 3, so we can rewrite the equation as $(3^{3})^{x+1}=3^{3x+3}=3^{2}$. From the uniqueness of $b^{x}$, we know that $b^{x}=b^{y}$ is equivalent to $x=y$ (when $b\gt0$ and $b\ne1$). Therefore, $3x+3=2$. Subtract 3 from both sides. $3x=-1$ Divide both sides by 3. $x=-\frac{1}{3}$