Ta có:
\(2^x+2^{x+1}+2^{x+2}=224\)
\(\Rightarrow2^x+2^x.2+2^x.2.2=224\)
\(\Rightarrow2^x\left(1+2+4\right)=224\)
\(\Rightarrow2^x=224\div\left(1+2+4\right)\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
Vậy \(x=5\)
2x + 2x + 1 + 2x + 2 = 224
2x + 2x . 2 + 2x . 22 = 224
2x( 1 + 2 + 4 ) = 224
2x . 7 = 224
\(\Rightarrow\)2x = 224 : 7
\(\Rightarrow\)2x = 32
\(\Rightarrow\)2x = 25
\(\Rightarrow\)x = 5
Vậy ..
2x + 2x+1 + 2x+2 =224
=>2x ( 1+2+22) =224
=> 2x . 7 =224
=> 2x =32
=>2x = 25
=>x =5
\(2^x+2^{x+1}+2^{x+2}=224\)
\(2^x+2^{x+1}+2^{x+2}=224\)\(2^x+2^{x+1}+2^{x+2}=224\)⇔ \(2^x+2^x.2+2^x.2^2=224\)
⇔ \(2^x\left(1+2+2^2\right)=224\)
⇔ \(2^x.7=224\)
⇔ \(2^x=224:7\)
⇔ \(2^x=32\)
⇔ \(2^x=2^5\)
⇒ \(x=5\)