Đặt ${{\log }_{5}}3=a$, khi đó ${{\log }_{9}}1125$ bằng

Ta có: ${{\log }_{9}}1125={{\log }_{{{3}^{2}}}}{{5}^{3}}{{.3}^{2}}={{\log }_{{{3}^{2}}}}{{5}^{3}}+{{\log }_{{{3}^{2}}}}{{3}^{2}}=\dfrac{3}{2}{{\log }_{3}}5+1=\dfrac{3}{2}\dfrac{1}{{{\log }_{5}}3}+1=1+\dfrac{3}{2a}$.