Question

Determinati numarul "n", pentru care are loc egalitatea 27^672 + 2x3^2016=3^4"n"-3.

Answer 1

(3³)^672+2·3^2016=3^(4n-3)
3^2016+2·3^2016=3^(4n-3)
3^2016(1+2)=3^(4n-3)
3^2017=3^(4n-3)
4n-3=2017
4n=2020
n=2020/4
n=505

Answer 2

27^672 + 2×3^2016=3^(4n-3)
(3^3)^672 + 2×3^2016=3^(4n-3)
3^2016+2×3^2016= 3^(4n-3)
3^2016 ×(1+2)=3^(4n-3)
3^2016×3=3^(4n-3)
3^2017=3^(4n-3)
2017=4n-3
4n=2020
n=2020:4=505