Question

Tìm x

1). 2^x+2 - 2^x+1 = 32

2). 3^x+5 + 3^x+2 = 756

Answer 1

1) Ta có: \(2^{x+2}-2^{x+1}=32\)

\(\Leftrightarrow2^{x+1}\left(2^1-1\right)=32\)

\(\Leftrightarrow2^{x+1}=2^5\)

\(\Leftrightarrow x+1=5\)

hay x=4

Vậy: x=4

2) Ta có: \(3^{x+5}+3^{x+2}=756\)

\(\Leftrightarrow3^{x+2}\left(3^3+1\right)=756\)

\(\Leftrightarrow3^{x+2}\cdot28=756\)

\(\Leftrightarrow3^{x+2}=27\)

\(\Leftrightarrow x+2=3\)

hay x=1

Vậy: x=1

Answer 2

1)

=> \(2^x.2^2-2^x.2=2^5\)

=> \(2^x.\left(2^2-2\right)=32\)

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

=> \(2^x=16\)

=> x=4

2)

=> \(3^x.3^5+3^x.3^2=756\)

=> \(3^x.\left(3^5+3^2\right)=756\)

=> \(3^x.252\)=756

=> \(3^x=3\)

=> x = 1