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

 \(2^{n+5}=2^{10}\) 

\(\Rightarrow n=5\) 

a) \(2^{x+1}-2^x=32\)

\(\Rightarrow2^x\left(2-1\right)=2^5\)

\(\Rightarrow2^x.1=2^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=2\)

\(\Rightarrow2^{x+1}=2^1\)

\(\Rightarrow x+1=1\)

\(\Rightarrow x=0\)

\(2^{x+1}-2^x=32\)

\(2^x.2-2^x=32\)

\(2^x\left(2-1\right)=32\)

\(2^x=32\)

\(x=5\)

\(2^{2x+1}=\left(2^2\right)^x.2=4^x.2\)

\(2^{x+1}-2^x=32\\ \Rightarrow2^x\cdot2^1-2^x=32\\ \Rightarrow2^x\left(2-1\right)=32\\ \Rightarrow2^x=32:1\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)

\(2^{x+1}-2^x=32\\ =>2^x.\left(2^1-1\right)=32\\ =>2^x.1=2^5\\ =>x=5\)

1. 2^x=16\(\Rightarrow\)X=4

2. 2^2x-1=2⁷

\(\Rightarrow\)2⁷=2^2.4-1

Vậy x =4

x=4 nha chị

\(2^x+2^{x+2}=32.\left(2^2+1\right)\)

\(\Rightarrow2^x+2^{x+2}=32.5\)

\(\Rightarrow2^x+2^{x+2}=160\)

\(\Rightarrow2^x\left(1+4\right)=160\)

\(\Rightarrow2^x.5=160\)

\(\Rightarrow2^x=160:5=32\)

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

\(\Rightarrow x=5\)

Vậy x = 5

               Bài làm :

Ta có :

\(x^2.x^3=32\Leftrightarrow x^{2+3}=32\Leftrightarrow x^5=2^5\Leftrightarrow x=2\)

Vậy x=2

\(x^2.x^3=32\)

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

x² . x³ = 32 
x² . x³ = 25
=> x =2 
Thử lại: 2² . 2³ = 2^{(2 + 3)} = 2⁵ 
 

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

\(2\times\left(4^2\right)^x=32\)

\(2\times16^x=32\)

\(16^x=16\)

\(x=1\)

\(2\left(4^2\right)^x=32\)

\(32^x=32\)

\(x=1\)