[tex]\it a=6^{15}=(2\cdot3)^{15}=2^{15}\cdot3^{15}\\ \\ b = 3^{25}\\ \\ \dfrac{a}{b}=\dfrac{2^{15}\cdot3^{15}^{(3^{15}}}{3^{25}} =\dfrac{2^{15}}{3^{10}}=\dfrac{2^{3\cdot5}}{3^{2\cdot5}}=\dfrac{(2^3)^5}{(3^2)^5}=\dfrac{8^5}{9^5}=\Big(\dfrac{8}{9}\Big)^5<1 \Rightarrow b>a[/tex]
