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

I am doing this exercise for the GMAT test

Which of the following option is closest to $$49^{4}81^{5}$$?

A. $$8^{18}$$

B. $$8^{19}$$

C. $$8^{20}$$

D. $$8^{21}$$

E. $$8^{22}$$

My attempt:

$$49^{4}81^{5} = (7^2)^4(9^2)^5 = 7^{8}9^{10} = (8-1)^{8}(8+1)^{10} \approx 8^{8}8^{10} = 8^{18}$$

But I am not sure how $$(8-1)^{8}(8+1)^{10}$$ is closer to $$8^{18}$$ than to $$8^{19}$$.

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 $$8^3$$.
Use the approximations $$\,49 \approx 50 = 10^2\cdot 2^{-1}\,$$ and $$\,81 \approx 80 = 10\cdot 2^3.\,$$ This gives us the approximation $$\, x := 49^4\cdot 81^5 \approx 10^8\cdot 2^{-4}\cdot 10^5\cdot 2^{15} = 10^{13}\cdot 2^{11}\,$$ and since $$\,10^3 \approx 2^{10}\,$$ we further get the approximation $$\, x \approx 10\cdot 2^{40}\cdot 2^{11} = 10\cdot 2^{51} = 10\cdot 8^{17} \approx 8^{18}.\,$$