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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 $8^{19}$.

Please shed me some lights. Thank you for your help!

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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 $8^3$.

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  • $\begingroup$ Thank you so much! I got it. $\endgroup$ – Le Anh Dung Mar 16 at 6:01
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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 $\, x \approx 10\cdot 2^{40}\cdot 2^{11} = 10\cdot 2^{51} = 10\cdot 8^{17} \approx 8^{18}.\,$

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