\(a,2^x+2^{x+1}=96\)
\(\Rightarrow2^x+2^x.2=96\) \(\Rightarrow2^x\left(1+2\right)=96\)
\(\Rightarrow2^x.3=96\) \(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(b,3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\) (Vô lý)
Vậy \(x\in\varnothing\)
a/ \(2^x+2^{x+1}=96\)
\(2^x+2^x.2=96\)
\(2^x\cdot\left(2+1\right)=96\)
\(2^x=\frac{96}{3}=32\)
\(2^x=2^5\)
\(=>x=5\)
b/ \(3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}-81^{x+3}=0\)
\(3^{4x}.3^4-3^{4x}\cdot81^3=0\)
\(3^{4x}\cdot\left(81-81^3\right)=0\)
\(3^{4x}=\frac{0}{81-81^3}\)
\(3^{4x}=0\Rightarrow x=0\)
a) 2x + 2x x 21= 96
2x x (1+ 2)=96
2x = 96 : 3= 32 =25
x= 5
