Answer
$y=\frac{3^{\ln_2{3.75}}}{4}\cdot x^{\ln_2{3.75}}$
Work Step by Step
Let $y=ax^b$.
Then our equations are: $4=a3^b$ and $15=a6^b$.
If we divide the second equation by the first one we get: $3.75=2^b$.
Then: $b=\ln_2{3.75}$.
Then $4=a3^{\ln_2{3.75}}\\a=\frac{3^{\ln_2{3.75}}}{4}$
Thus $y=\frac{3^{\ln_2{3.75}}}{4}\cdot x^{\ln_2{3.75}}$