## Precalculus (10th Edition)

RECALL: (1) $\dfrac{1}{a^x} = a^{-x}$ (2) $(a^m)^n=a^{mn}$ Use rule (1) above to obtain: $\left(\dfrac{1}{3}\right)^x = \left(3^{-1}\right)^x$ Use rule (2) above to obtain: $\left(3^{-1}\right)^x = 3^{-1(x)} = 3^{-x}$ Note that $3^x \ne 3^{-x}$, so $y=3^x$ is not equivalent to $y=\left(\dfrac{1}{3}\right)^x$. This means that they have different graphs. Thus, the given statement is false.