a: Ta có: \(3^{2x+1}< 27\)

\(\Leftrightarrow2x+1< 3\)

\(\Leftrightarrow x< 1\)

hay x=0

1. 

a. 3^{2x + 1} < 27

<=> 3^{2x + 1} < 3³

<=> 2x + 1 < 3

<=> 2x < 2

<=> 2x : 2 < 2 : 2

<=> x < 1

\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)

Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)

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

Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)

cứu

a: 16^19=(2^4)^19=2^76

8^25=(2^3)^25=2^75

mà 76>75

nên 16^19>8^25

b: 3^500=(3^5)^100=243^100>5^100

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

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`.

b)

a = 25.26 261 = 25.(26 260 +1) = 25.10.2626 + 25 = 25.10.26.101 + 25

b = 26.25 251 = 26.(25 250 + 1) = 26.10.2525 + 26 = 26.10.25.101 + 26

Suy ra a < b

a=25.26261=25.(26260+1) = 25.10.2626+25 = 25.10.26.101+25

b=26.25251=26.(25 250+1)=26.10.2525+26=26.10.25.101+26

Vì 26>25 nên b>a

a<b vì:

25<26

bài nào

bài nào bn

???

a)2^6=8^2

b)5^3<3^5

a) \(2^6\) và \(8^2\)

   \(2^6=\left(2^2\right)^3\)

  \(8^2=\left(2^3\right)^2\)\(=2^6\)

 \(\Rightarrow\) \(2^6=8^2\)

a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

a) 0,(26)<0,261

b) 0,15>0,14(9)

a: 0,(26)<0,261

b: 0,15>0,14(9)