How to approximate $49^{4}81^{5}$?

Let's take it exactly for a few more steps: $$ (8-1)^{8}(8+1)^{10} =(8-1)^8(8+1)^8(8+1)^2\\ =(8^2-1)^8\cdot9^2<8^{16}\cdot 9^2 $$ And $9^2$ is much closer to $8^2$ than to $8^3$.