## Precalculus (6th Edition) Blitzer

$y=6.5 e^{-0.844x}$
Recall the log rules: (a) $\log_b{\dfrac{p}{q}}=\log_b{p} - \log_b{q}$ (Quotient Rule). (b) $\log{a}=\log_{10}{a}$ (c) $\log_a{a^x}=x$ (d) $e^{\ln a}=a$ We are given $y=6.5(0.43)^x$ ....(1) We simplify equation (1) by using rule (d) with $a=0.43$: $y= 6.5e^{\ln 0.43 x}=6.5 e^{-0.844x}$