`@` `\text {Ans}`

`\downarrow`

`a.`

`3^n = 27` phải k c?

`3^n = 27`

`=> 3^n = 3^3`

`=> n=3`

Vậy, `n=3`

TH2 (đề):

`3n = 27`

`=> n = 27 \div 3`

`=> n=9`

Vậy, `n=9`

`b.`

TH1:

`5^n = 625`

`=> 5^n = 5^4`

`=> n = 4`

Vậy, `n=4`

TH2: 

`5n = 625`

`=> n = 625 \div 5`

`=> n = 125`

Vậy, `n=125`

Bài 1:

a) \(\frac{\left(-3\right)^n}{81}=9\Leftrightarrow\left(-3\right)^n=9.81=729\Rightarrow\left(-3\right)^n=\left(-3\right)^6\Rightarrow n=6\)

b) \(\frac{125}{5^n}=5^2\Leftrightarrow\frac{125}{5^n}=25\Rightarrow5^n=125:25=5\Rightarrow n=1\)

Bài 2:

a) \(625^5=\left(5^4\right)^5=5^{4.5}=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{3.7}=5^{21}\)

Thấy: \(5^{20}< 5^{21}\Rightarrow625^5< 125^7\)

b) \(3^{2n}=\left(3^2\right)^n=9^n\) ; \(2^{3n}=\left(2^3\right)^n=8^n\)

\(9^n>8^n\Rightarrow3^{2n}>2^{3n}\)

K cho mình nhé.

a/  n= 27 : 3 = 9

b/ n = 81 : 3  = 27

c/   5n = 125 -2

<=> 5n = 123

<=>  n =\(\frac{123}{5}\)

d/  ( n -1 ) ⁴ = 64

<=> ( n - 1)² = 8

<=> n² -2n + 1 = 8

<=> n² -2n -7 = 0

<=> n = \(1+2\sqrt{2}\)

     n= \(1-2\sqrt{2}\)

OK

a) \(3n=27\Rightarrow n=27:3=9\)

b) \(81=3n\Rightarrow n=81:3=27\)

c) \(5n+2=125\Rightarrow5n=125-2=123\Rightarrow n=\frac{123}{5}\)

d) \(\)n thuộc rỗng vì không có số nào có mũ 4=64 nha

1. a) 625/5n=5³ => 5n=625/5³=5⁴/5³=5 =>n=1

b) (-2n)/-128=4 =>-2n=4.(-128)=-2.256 =>n=256

c) (3/7)ⁿ=81/2401=(3/7)⁴ => n=4

2. 32<2ⁿ<512

<=> 2⁵<2ⁿ<2⁹

=> n=6;7;8

3. (x-1)⁴=16=2⁴ => x-1=2 =>x=3

2^x = 32

=> 2^x = 2⁵

=> x = 5

vậy_

a) \(2^x=32\)

Ta có: \(2^5=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

b) Sửa đề tí: \(9< 3^x< 81\)

\(\Rightarrow3^2< 3^x< 3^4\)

\(\Rightarrow2< x< 4\)

\(\Rightarrow x=\left\{3\right\}\)

Vậy x = 3

c) Ta có: \(25\le5^x\le125\)

\(\Rightarrow5^2\le5^x\le5^3\)

\(\Rightarrow2\le x\le3\)

\(\Rightarrow x=\left\{2;3\right\}\)

Vậy x = 2 hoặc x = 3

d) \(\left(x-2\right)^3\times5=40\)

\(\Rightarrow\left(x-2\right)^3=8\)

Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)

Suy ra: x - 2 = 2

Vậy x = 4

\(3^x>9< 81\)

\(=>3^x>3^2< 3^4\)

\(=>3^x=3^3\)

\(=>x=3\)

a, ta co :

x ∈ B[9] = { 0 ; 9 ; 18 ; 27 ; 36 ;45 ;54 ;63 ;72 ; 81 ;90 ;...............}

vi 45 ≤ x ≤ 81

Suy ra : x ∈ { 45 ; 54 ; 63 ; 72; 81 }

b, |x+3| = 9

ta co 2 truong hop :

trường hợp 1 : x + 3 = -9 trường hợp 2 : x + 3 = 9

x = -9 - 3 x = 9 - 3

x = -12 x = 6

x ∈ B[9] = { 0 ; 9 ; 18 ; 27 ; 36 ;45 ;54 ;63 ;72 ; 81 ;90 ;...............}

vi 45 ≤ x ≤ 81

Suy ra : x ∈ { 45 ; 54 ; 63 ; 72; 81 }

b, |x+3| = 9

ta co 2 truong hop :

trường hợp 1 : x + 3 = -9

x = -9 - 3

x = -12

trường hợp 2 : x + 3 = 9

x = 9 - 3

x = 6

vay x = 6 ; x = -12

b)/x + 3/ = 9

x = 6

x= -12

1.

a. 9 < \(^{3^y}\)< 81

9=3\(^{^2}\)

81=3\(^{^4}\)

\(\Rightarrow\)y=3 vì \(3^2< 3^3< 3^4\)

b.25\(\le\)5\(^{^y}\)\(\le\)125

25=5\(^{^2}\)

125=5\(^{^3}\)

\(\Rightarrow\)y=2 và 3 vì 5\(^{^2}\)\(\le\)5\(^{^2}\)(5\(^{^3}\))\(\le\)5\(^{^3}\)

c.

16\(\ge\)4\(^{^y}\)\(\ge\)1024

16=4\(^{^2}\)

1024=4\(^{^5}\)

\(\Rightarrow\)y=2,3,4,5 vì 4​​\(^{^2}\)​​​\(\ge\)4\(^{^2\left(4^3;4^4;4^5\right)}\)

\(^{^2}\)

\(\Rightarrow\)

2.

a.3x\(^{2^y}\)=48

\(^{2^y}\)=48:3

\(^{2^y}\)=16

\(\Rightarrow\)y=4 vì \(^{2^4}\)=16

b.5x\(y^7\)=640

\(y^7\)=640:5

\(y^7\)=128

\(\Rightarrow\)y=2 vì \(2^7\)=128

c.\(y^{100}\)=\(y^2\)

\(\Rightarrow\)y=1 vì:

\(1^{100}\)=1

\(1^2\)=1

d.(y-3)\(^{^5}\)=243

\(\Rightarrow\)y-3=3 vì 3\(^{^5}\)=243

y=3+3

y=6

e.(2y+1)\(^{^3}\)=125

\(\Rightarrow\)2y+1=5 vì 5\(^{^3}\)=125

2y=5-1=4

y=4:2=2

i.2\(^{^{y+3}}\)+2\(^{^y}\)=288

2\(^{^y}\).2\(^{^3}\)+2\(^{^y}\)=288

2\(^{^y}\).2\(^{^3}\)=288-2\(^{^y}\)

2\(^{^y}\).8=288-2\(^{^y}\)

8=(288-2\(^{^y}\)):2\(^{^y}\)

8=288:2\(^{^y}\)-2\(^{^y}\):2\(^{^y}\)

8=288:2\(^{^y}\)-1

288:2\(^{^y}\)=8+1=9

2\(^{^y}\)=288:9=32

\(\Rightarrow\)y=5 vì 2\(^{^5}\)=32

\(a,\)

Ta có : 

\(\hept{\begin{cases}x\le y\\y\le x\end{cases}}\Rightarrow x=y\)

Mà \(y\le5\le x\)

\(\Rightarrow x=y=5\)

\(b,\)

Ta có : \(\hept{\begin{cases}b\le a\\a\le3\end{cases}}\Rightarrow b\le a\le3\)

Mà \(a\le b\)

\(\Leftrightarrow a=b\)

\(a,b\in N\Rightarrow a,b=\left\{0;1;2;3\right\}\)

a, 

Ta có :

    \(x\le y\)

   \(y\le5\le x\)

Ta có : 

  \(x\le y\) \(;\) \(y\le x\)

\(\Leftrightarrow x=y\)

\(y\le5\Leftrightarrow x;y\le5\)

Mà \(x;y\in N\)

\(\Leftrightarrow x;y\in\left\{0;1;2;3;4;5\right\}\)

+,Còn ý b tương tự bạn tự làm nha! dễ mà.