Question

Tìm x: \(4^x+5.2^{2x+1}=176\)

Answer 1

\(4^x+5.2^{2x+1}=176\)

\(4^x+5.4^x.2=176\)

\(4^x.\left(1+10\right)=176\)

\(4^x=176:11\)

\(4^x=16\)

\(4^x=4^2\)

\(=>x=2\)

Answer 2

\(4^x+5.2^{2x+1}=176\)
\(\Rightarrow2^{2x}+5.2^{2x}.2=176\)
\(\Rightarrow2^{2x}\left(1+10\right)=176\)
\(\Rightarrow2^{2x}.11=176\)
\(\Rightarrow2^{2x}=16\)
\(\Rightarrow2^{2x}=2^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)

Answer 3

\(4^x+5.2^{2x+1}=176\\ 4^x+5.2^{2x}.2=176\\ 4^x+5.4^x.2\\ 4^x+4^x.10=176\\ 4^x.\left(10+1\right)=176\\ 4^x=176:11\\ 4^x=16\\ 4^x=4^2\\ x=2\)

Answer 4