## Algebra 2 (1st Edition)

$y=313\cdot\sqrt2^x$
Let $y=ab^x$. Then two of our equations are: $313=ab^0$ and $616=ab^2$. If we divide the second equation by the first one we get: $2=b^2$, thus because $b$ is positive: $b=\sqrt2$. Also $b^0=1$, thus $a=313$. Thus $y=313\cdot\sqrt2^x$