Question

comparati nr:
a)3^2^3 cu (3^2)^3(este 3 la 2 si apoi un 3 mai sus de 2 la inceput)
b) 3^38 cu 2^59-2^59-2^57
c)9^40 cu 81^20
pls repede imi trebuie acum va rogg​

Answer 1

a) 3^(2^3) = 3^8

(3^2)^3 = 3^6

3^8 > 3^6 => 3^(2^3) > (3^2)^3

b) 3^38 = (3^2)^19 = 9^18

2^59 - 2^58 - 2^57 = 2^57 × (2^2 - 2 - 1) = 2^57 × 1 = 2^57 = (2^3)^18 = 8^18

9^18 > 8^18 => 3^38 > 2^59 - 2^58 - 2^57

c) 9^40 = (3^2)^40 = 3^80

81^20 = (3^4)^20 = 3^80

9^40 = 81^20

Answer 2

Răspuns:

3²^³=3⁸

(3²)³=3⁶

3⁸ > 3⁶

3²^³ > (3²)³

___________

3³⁸=(3²)¹⁹=9¹⁸

2⁵⁹-2⁵⁸-2⁵⁷=2⁵⁷·(2²-2-1)=2⁵⁷·1=2⁵⁷=(2³)¹⁸=8¹⁸

9¹⁸ > 8¹⁸

3³⁸ > 2⁵⁹-2⁵⁸-2⁵⁷

_____________

9⁴⁰=(3²)⁴⁰=3⁸⁰

81²⁰=(3⁴)²⁰=3⁸⁰

3⁸⁰=3⁸⁰

9⁴⁰=81²⁰