a: \(4^{x+2}-2^{2x}=60\)
=>\(4^x\cdot16-4^x=60\)
=>\(4^x\cdot\left(16-1\right)=60\)
=>\(4^x=4\)
=>x=1
\(a,4^{x+2}-2^{2x}=60\)
\(\Rightarrow4^x.4^2-4^x=60\)
\(\Rightarrow4^x\left(4^2-1\right)=60\)
\(\Rightarrow4^x\left(16-1\right)=60\)
\(\Rightarrow4^x.15=60\)
\(\Rightarrow4^x=60:15\)
\(\Rightarrow4^x=4\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b,\(10.2^{x-1}+2^{x+2}\)
\(\Rightarrow10.2^{x-1}+2^{x-1+3}\)
\(\Rightarrow10.2^{x-1}+2^{x-1}.2^3\)
\(\Rightarrow2^{x-1}.\left(10+2^3\right)\)
\(\Rightarrow2^{x-1}.\left(10+8\right)\)
\(\Rightarrow2^{x-1}.18\)
\(\Rightarrow2^{x-1}.2.9\)
\(\Rightarrow2.2^{x-1}.9\)
\(\Rightarrow2^{1+x-1}.9\)
\(\Rightarrow2^x.9\)