## Algebra and Trigonometry 10th Edition

$y= e^{\frac{x}{3}\ln \frac{1}{4}}$
We have $y=ae^{bx}$ Set $(x,y)= (0,1)$ to compute $a$ So, $1=ae^{b(0)} \implies a=1$ Further, we have $y=ae^{bx}$ Set $(x,y)= (3,\dfrac{1}{4})$ to compute $b$ So, $\dfrac{1}{4}=ae^{3b} \implies \dfrac{1}{4}=e^{3b}$ and $3b =\ln \dfrac{1}{4} \implies b =\dfrac{1}{3} \ln \dfrac{1}{4}$ Now, the exponential decay model is: $y= e^{\frac{x}{3}\ln \frac{1}{4}}$