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pls

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\(2^{2n}+1=32\)

\(\Rightarrow2^{2n}+1=2^5\)

\(\Rightarrow2^{2n}+1=5\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=2\)

a,     2^n =32

<=> 2^n  = 2^5

=>       n=5

\(8^n:2^n=4\)

\(\Leftrightarrow\left(2^3\right)^n:2^n=2^2\)

\(\Leftrightarrow2^{3n}:2^n=2^2\)

\(\Leftrightarrow3n-n=2\)

\(\Leftrightarrow2n=2\)

\(\Leftrightarrow n=1\)

a) \(\frac{16}{2^n}=2\)

=> 2.2ⁿ = 16

=> 2¹⁺ⁿ = 2⁴

=> 1 + n = 4

=> n = 4 - 1

=> n = 3

Vậy n = 3

b) \(\frac{\left(-3\right)^n}{81}=-27\)

=> (-3)ⁿ = -27.81

=> (-3)ⁿ = -3³.3⁴

=> (-3)ⁿ = (-3)⁷

=> n = 7

Vậy n = 7

c) 8ⁿ : 2ⁿ = 4

=> (8 : 2)ⁿ = 4

=> 4ⁿ = 4¹

=> n = 1

Vậy n = 1

a ) 2^x = 32 
=> 2⁵       
b ) 3^x = 243
=> 3⁵
c ) 2^x = 256
=> 2⁸ 

a,x=5

b,x=5

c,x=8

TL:

a.\(2^6.2^n=2^{11}\) 

 \(2^{6+n}=2^{11}\) 

\(\Rightarrow n=5\) 

b. \(3^7:3^n=3^4\) 

  \(3^{7-n}=3^4\) 

\(\Rightarrow n=3\) 

c.\(2^n.32=2^{10}\)

 \(2^{n+5}=2^{10}\) 

\(\Rightarrow n=5\) 

Ta có :

13²ⁿ = 169

=> 13²ⁿ = 13²

=> 2n = 1

=> n = 1

( n - 4 )⁵ = 32

=> ( n - 4 )⁵ = 2⁵

=> n - 4 = 2

=> n = 6

64 . 4^x = 168

<=> 43. 4^x = 416

=> 3 + x = 16

<=> x = 13

Vậy x = 13

2^x.16² = 1024

<=> 2^x. 2⁸ = 210

=> x + 8 = 10

<=> x = 2

Vậy x = 2

b: Ta có: \(2^x\cdot16^2=1024\)

\(\Leftrightarrow2^x\cdot2^8=2^{10}\)

\(\Leftrightarrow x+8=10\)

hay x=2

  64.4^x=16³

  2⁶. (2²)^x = (2⁴)³

=> 2⁶. 2^2x = 2¹²

        2^2x   = 2 ¹² : 2⁶

       2^2x   = 2⁶

  =>  2x = 6

         x = 6 :2

         x=3

- Vậy x =3