\(a) 3^{200}=(3^2)^{100}=9^{100}\\2^{300}=(2^3)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\) nên \(3^{200}>2^{300}\)

\(b) 5^{40}=(5^4)^{10}=625^{10}\\3^{50}=(3^5)^{10}=243^{10}\)

Vì \(625^{10}>243^{10}\) nên \(5^{40}>3^{50}\)

#\(Toru\)

a> \(3^{200}\) và \(2^{300}\)

Ta có:\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

          \(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

Vì 9>8 nên \(9^{100}>8^{100}\)

\(\Rightarrow\)\(3^{200}>2^{300}\)

b> \(5^{40}\) và \(3^{50}\)

Ta có:\(5^{40}=5^{4.10}=\left(5^4\right)^{10}=625^{10}\)

         \(3^{50}=3^{5.10}=\left(3^5\right)^{10}=243^{10}\)

Vì 625 > 243 nên \(625^{10}>243^{10}\)

\(\Rightarrow\)\(5^{40}>3^{50}\)

`3^200=(3^2)^100=9^100`.

`2^300=(2^3)^100=8^100`.

`=> 2^300 < 3^200`.

`b, 5^40=(5^4)^10=625^10.`

`3^50=(3^5)^10=243^10`.

`=> 5^40 > 3^50`.

a) >

b) <

c) <

d) <

a)>

b)<

c)<

d)<

a) Cách 1:  2 100 = 2 10 10 = 1024 10 > 1024 9

Cách 2:  1024 9 = 2 10 9 =  2 90  <  2 100

b)  6 . 5 29  >  5 . 5 29  =  5 30

c)  2 98 =  2 2 49 =  4 49  <  9 49

d)  10 30 =  10 3 10 = 1000 10 ;  2 100 =  2 10 10  = 1024 10 nên  10 30  <  2 100

a)  1024 9 = ( 2 10 ) 9 = 2 90 < 2 100

b)  6 . 5 29 > 5 . 5 29 = 5 30

c) 10 30 = ( 10 3 ) 10 = 1000 10 ; 2 100 = ( 2 10 ) 10 = 1024 10   n ê n   10 30 < 2 100 .

a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

c) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

Giải chi tiết giúp mình ạ~

\(\left(d\right):202^{303}=\left(202^3\right)^{101}=8242408^{101}>303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(\left(e\right):107^{50}=\left(107^2\right)^{25}=11449^{25}< 73^{75}=\left(73^3\right)^{25}=389017^{25}\)

a) \(2^x=16=2^4\Rightarrow x=4\)

b) \(x^3=27=3^3\Rightarrow x=3\)

c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)

\(\Rightarrow x=6\) hay \(x=-2\)

a) \(2^{300}=2^{3.100}=8^{100}\)

\(3^{200}=3^{2.100}=9^{100}\)

vì \(8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(3^{500}=3^{5.100}=243^{100}\)

\(7^{300}=7^{3.100}=343^{100}\)

vì \(243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a,`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4`

`b,`

`x^3 = 27`

`=> x^3 = 3^3`

`=> x = 3`

Vậy, `x = 3`

`c,`

\(x^{50}=x\)

`=>`\(x^{50}-x=0\)

`=>`\(x\left(x^{49}-1\right)=0\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy, `x \in {0; 1}`

`d,`

`(x-2^2)=16`

`=> x - 2^2 = 16`

`=> x = 16 + 2^2`

`=> x = 20`

Vậy, `x = 20`

`2,`

`a,`

Ta có:

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì `8 < 9 =>`\(8^{100}< 9^{100}\)

`=>`\(2^{300}< 3^{200}\)

Vậy, \(2^{300}< 3^{200}\)

`b,`

Ta có:

\(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}\)

Vì `243 < 343 =>`\(243^{100}< 343^{100}\)

`=>`\(3^{500}< 7^{300}\)

Vậy, \(3^{500}< 7^{300}.\)

a: \(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

mà 8<9

nên \(2^{300}< 3^{200}\)

b: \(3^{500}=243^{100}\)

\(7^{300}=343^{100}\)

mà 243<243

nên \(3^{500}< 7^{300}\)

2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰ và 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰ nên 2³⁰⁰ < 3²⁰⁰;