\(\text{a) Ta co }\) \(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(\Rightarrow\) \(4^{x+1}\left(16-3\right)=13.4^{11}\)
\(\Rightarrow4^{x+1}.13=13.4^{11}\)
\(\Rightarrow4^{x+1}=4^{11}\)
\(\Rightarrow x+1=11\)
\(\Rightarrow\text{x=10}\)
a)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
<=> \(4^{x+1}\left(16-3\right)=13.4^{11}\)
<=> \(4^{x+1}.13=13.4^{11}\)
<=> \(4^{x+1}=4^{11}\)
<=> \(x+1=11\)
<=> x=10