Se $11,25 = \Large{\frac{1125}{100}}$ $=$ $\Large{\frac{5^3\cdot 3^2}{10^2}}$ $=$ $3^2 \cdot 10 \cdot 2^{-3}$, então:$$\log_{15} \sqrt[5] {3^2 \cdot 10 \cdot 2^{-3}} = \frac{\log 3^2 + \log 10 + \log 2^{-3}}{5\cdot \log 15} = \boxed{\frac{2b + 1 -3a}{5\cdot( b+1-a)}}$$ Obs: $\log 15 = \log 3 + \log 10 - \log 2 = (b+1-a)$