## Algebra 2 (1st Edition)

Let $y=ax^b$. Then two of our equations are: $0.5=a5^b$ and $0.75=a11^b$. If we divide the second equation by the first one we get: $1.5=2.2^b=\\b=\ln_{2.2}{1.5}\approx0.514$. Then $0.5=a5^{0.514}\\a=\frac{0.5}{5^{0.514}}\approx0.219$. Thus $y=0.219\cdot x^{0.514}$ Then plugging in $x=120$ we get: $y=0.219\cdot (120)^{0.514}\approx2.5$ is required.