smc

We want to find $r$, so our strategy is to eliminate $k$. The first equation gives $k=\frac{3}{2^r}$. The second equation gives $k=\frac{15}{4^r}$ Setting those two equal gives $\frac{3}{2^r}=\frac{15}{4^r}$ Cross-multiplying and dividing by $3$ gives $5\cdot 2^r=4^r$. We know that $4^r=2^r\cdot 2^r$, so we can divide out $2^r$ from both sides (which is legal since $2^r\ne 0$), and we get: $5=2^r$ $r={\log }_{2}5$, which is option $\boxed{{\log }_{2}5}$.