Question

Va rog sa Ma ajutați cant mai repede posibil

Answer 1

Răspuns:

Explicație pas cu pas:

8^17 = (2^3)^17 = 2^51

16^12 = (2^4)^12 = 2^48

8^17 > 16^12

____________

27^13 = (3^3)^13 = 3^39

9^20 = (3^2)^20 = 3^40

27^13 < 9^20

____________

9^13 = (3^2)^13 = 3^26

3^27 > 9^13

____________

2^33 = (2^3)^11 = 8^11

3^22 = (3^2)^11 = 9^11

2^33 < 3^22

Answer 2

Răspuns:

Explicație pas cu pas:

b)

$8^{17} = (2^3)^{17} = 2^{3*17} = 2^{51}\\16^{12} =(2^4)^{12}= 2^{4*12} = 2^{48} < 2^{51}\\\\8^{17} > 16^{12}$

c)

$3^{39}= (3^3)^{13} = 3^{3*13} = 3^{39}\\9^{20} =(3^2)^{20}=3^{2*20} = 3^{40} > 3^{39}\\27^{13} < 9^{20}$

d)

$9^{13} =(3^2)^{13}=3^{2*13} = 3^{26} < 3^{27}\\3^{27} > 9^{13}$

e) fie numerele x si y asa incat:

$2^{33} = 10^x$

$3^{22} = 10^y$

Avem:

$2^{33} = 10^x \\log(2^{33}) = log(10^x)\\33*log(2) = x*log(10)\\33*log(2) = x\\x=33*log(2)=33*0,3010=9,9339$

Si

$3^{22} = 10^y \\log(3^{22}) = log(10^y)\\22*log(3) = y*log(10)\\22*log(2) = y\\y=22*log(3)=22*0,4771=10,4966$

$10,4996 > 9,9339\\y > x\\10^y>10^x\\3^{22} > 2^ {33}$