Ta có: \(2^x+2^{x+2}=20\)
\(\Rightarrow\) \(2^x+2^x.4=20\)
\(\Rightarrow\) \(2^x\left(1+4\right)=20\)
\(\Rightarrow2^x.5=20\)
\(\Rightarrow2^x=20:5\)
\(\Rightarrow\) \(2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
\(2^x+2^{x+2}=20\)
\(\Rightarrow2^x+2^x\times2^2=20\)
\(\Rightarrow2^x+2^x\times4=20\)
\(\Rightarrow2^x\times\left(1+4\right)=20\)
\(\Rightarrow2^x\times5=20\)
\(\Rightarrow2^x=20\div5\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\) thì thỏa mãn đề bài.
\(2^x+2^{x+2}=20\)
\(\Rightarrow2^x\left(1+2^2\right)=20\)
\(\Rightarrow2^x.5=20\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy....
Ta có : \(2^x+2^{x+2}=2^2+2^{2
+2}\)
Suy ra x=2
