## Precalculus (6th Edition) Blitzer

$y= 73 e^{0.956x}$
We are given: $y=73 (2.6)^x$ We use the fact that $e^{\ln{a}}=a$, with $a=2.6$: $y= 73 e^{(\ln 2.6) x}$ $y=73 e^{0.956x}$