64⁸=(2⁶)⁸=2⁴⁸

16¹²=(2⁴)¹²=2⁴⁸

Vì 2⁴⁸=2⁴⁸ nên 64⁸=16¹²

64⁸ = (2⁶)⁸ = 2⁴⁸

16^{12 =} (2⁴)¹² = 2⁴⁸

Vì 2⁴⁸ = 2⁴⁸ => 16¹² = 64⁸

Vậy 64⁸ = 16¹²

Trả lời

4¹⁰⁰=2²⁰⁰

2²⁰²

Vậy 2²⁰⁰ < 2²⁰² hay 4¹⁰⁰ < 2²⁰²

3⁰ và 5⁸

3⁰ < 5⁸

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d, 

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d,

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

6 mũ 37>16 mũ 12

Hoặc 6 mũ 37<16 mũ 12

Hoặc 6 mũ 37=16 mũ 12 

6 mũ 37 > 16 MŨ 12

a) 5^23 và 6 . 5^22

Ta có: 5^23 = 5^22 . 5

Vì 5 < 6 nên 5^23 < 6 . 5^22

b) 7 . 2^13 và 2^16

Ta có: 2^16 = 2^13 . 2^3 = 2^13 . 8

Vì 7 < 8 nên 7 . 2^13 < 2^16

c) 21^15 và 27^5 . 49^8

Ta có: 21^15 = (3.7)^15 = 3^15 . 7^15

27^5 . 49^8 = (3^3)^5 . (7^2)^8 = 3^15 . 7^16

Vì 7^15 < 7^16 nên 21^15 < 27^5 . 49^8

f

So sánh: 19²⁰ và 9⁸.5¹⁶

9⁸.5¹⁶=(3²)⁸.5¹⁶=3¹⁶.5¹⁶=(3.5)¹⁶=15¹⁶

Vì 19>15 và 20>16

Nên 19²⁰ > 9⁸.5¹⁶

19 mũ 20=19+20=39

9 mũ 8 nhân 5 mũ 16=9+8 nhân 5 +16=65

39<65

64² = (2⁶)² = 2¹²

16¹² = (2⁴)¹² = 2⁴⁸

Vì 2¹² < 2⁴⁸ => 64² < 16¹²

Vậy 64² < 16¹²

2²⁵>16⁶

\(16^6=\left(2^3\right)^6=2^{24}\)

Vì \(2^{24}< 2^{25}\)nên \(2^{25}>16^6\)

\(-\left(-\dfrac{1}{16}\right)^{100}=-\left(-\dfrac{1}{2^4}\right)^{100}=-\left(\dfrac{1}{2^4}\right)^{100}=-\left[\left(\dfrac{1}{2}\right)^4\right]^{100}=-\left(\dfrac{1}{2}\right)^{400}=-\dfrac{1}{2^{400}}\)

\(-\left(-\dfrac{1}{8}\right)^{150}=-\left(-\dfrac{1}{2^3}\right)^{150}=-\left(\dfrac{1}{2^3}\right)^{150}=-\left[\left(\dfrac{1}{2}\right)^3\right]^{150}=-\left(\dfrac{1}{2}\right)^{450}=-\dfrac{1}{2^{450}}\)

\(\dfrac{1}{2^{400}}>\dfrac{1}{2^{450}}\Rightarrow-\dfrac{1}{2^{400}}< -\dfrac{1}{2^{450}}\)

Vậy \(-\left(-\dfrac{1}{6}\right)^{100}< -\left(-\dfrac{1}{8}\right)^{150}\)