Ta có: 2¹⁶¹>2¹⁶⁰=2^4.40=(2⁴)⁴⁰=16⁴⁰>13⁴⁰

=>2¹⁶¹>13⁴⁰

2^161=(2^4)^40*2=16^40*2

ta thấy 16^40>13^40

suy ra 16^40*2>13^40

suy ra 2^161>13^40

2¹⁶¹ > 2¹⁶⁰ = 2^4.40 = (2⁴)⁴⁰ = 16⁴⁰ > 13⁴⁰ nên 2¹⁶¹ > 13⁴⁰.

Ta giữ nguyên 13⁴⁰

Ta thấy 2¹⁶¹>2¹⁶⁰

.Mà 2¹⁶⁰=2^4.40=(2⁴)⁴⁰=16⁴⁰ Do 16>13 nên 13⁴⁰<2¹⁶¹

Bài 1: 

a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)

b) \(x^{10}=x\Rightarrow x=1\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)

\(\left(2x-15\right)^2=1\Rightarrow x=8\)

Bài 2:

\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)

\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)

C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)

             \(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)

Vì 2⁵<10⁵ nên 199²⁰<2003¹⁵

\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)

Câu a:

10^30 và 2^100

(10^3)^10 = 1000^10

2^100 = (2^10)^10 = 1024^10 > 1000^10

Vậy 10^30 < 2^100

333^444 = (333^4)^111

444^333 = (444^3)^111

333^4 = 3^4.111^4 = 81.111^4

444^3 = 4^3.111^3 = 64.111^3 < 81.111^4

444^3< 333^4

(333^4)^111 > (444^3)^111

333^444 > 444^333

a) 125⁸⁰ = (5³)⁸⁰ = 5²⁴⁰

25¹¹⁸ = (5²)¹¹⁸ = 5²³⁶

Do: 240 > 236  => 5²⁴⁰ > 5²³⁶ hay 125⁸⁰ > 25¹¹⁸

b) 2¹⁶¹ = 2¹⁶⁰ x 2 = (2⁴)⁴⁰ x 2 = 16⁴⁰ x 2

Do: 16 > 13 nên 16⁴⁰ > 13⁴⁰ và 16⁴⁰ x 2 > 13⁴⁰

=> 13⁴⁰ < 2¹⁶¹

 a) 125⁸⁰ và 25¹¹⁸

125⁸⁰ = ( 5³) ⁸⁰= 5²⁴⁰

25¹¹⁸= ( 5² ) ¹¹⁸ = 5²³⁶

VÌ 5²⁴⁰ < 5²³⁶ nên = > 125⁸⁰ < 25¹¹⁸

b) 13⁴⁰ và 2¹⁶¹

2¹⁶¹ = ( 2⁴)¹⁵⁷ = 16¹⁵⁷

vì 13 < 16 nên 13⁴⁰ < 16¹⁵⁷

vậy 13⁴⁰ < 2¹⁶¹

Ta có :

\(2^{161}=\left(2^4\right)^{40}.2=16^{40}.2\)

Vì \(16^{40}.2>13^{40}\)

\(\Rightarrow2^{161}>13^{40}\)

Ta có: \(2^{161}>2^{160}=\left(2^4\right)^{40}=16^{40}\)

Vì \(16^{40}>13^{40}\) nên \(2^{160}>13^{40}\Rightarrow2^{161}>13^{40}\)