\(2^{x+1}-2^x=32\)
\(\Leftrightarrow2^x.2-2^x=2^5\)
\(\Leftrightarrow2^x.\left(2-1\right)=2^5\)
\(\Leftrightarrow2^x.1=2^5\)
\(\Leftrightarrow x=5\)
Vậy : \(x=5\)
\(2^{x+1}-2^x=32\)
\(\Leftrightarrow2^x.2-2^x=2^5\)
\(\Leftrightarrow2^x.\left(2-1\right)=2^5\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2x+1 -2x =32
2x.2 - 2x .1 =25
2x . (2 -1) =25
2x . 1 =25
Vay x =5