\(64\cdot4^x=16^2\)
=>\(4^x\cdot4^3=4^4\)
=>x+3=4
=>x=4-3=1
\(64.4^x=16^2\)
\(\Rightarrow4^3.4^x=\left(4^2\right)^2\)
\(\Rightarrow4^{3+x}=4^4\)
\(\Rightarrow3+x=4\)
\(\Rightarrow x=4-3\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
\(4^3+4^X=16^2 \)
\(4^{3+x}=4^4\)
3+x = 4
x=4-3
x=1