a)Ta có:
\(3^x-3^{x-3}=-234\)
\(\Rightarrow3^x-3^x\cdot3^3=-234\)
\(\Rightarrow3^x\cdot\left(1-3^3\right)=-234\)
\(\Rightarrow3^x\cdot\left(-26\right)=-234\)
\(\Rightarrow3^x=9\)
\(\Rightarrow x=2\)
Vậy x=2
\(\Rightarrow3^x=3^2\)
b) Ta có:
\(2^{x+1}\cdot3^x-6^x=216\)
\(\Rightarrow2^x\cdot2\cdot3^x-2^x\cdot3^x=216\)
\(\Rightarrow\left(2^x\cdot3^x\right)\cdot\left(2-1\right)=216\)
\(\Rightarrow6^x\cdot1=216\)
\(\Rightarrow6^x=6^3\)
\(\Rightarrow x=3\)
Vậy x=3
c) Ta có:
\(2^{2x+1}+4^{x+3}=264\)
\(\Rightarrow2^{2x+1}+\left(2^2\right)^{x+3}=264\)
\(\Rightarrow2^{2x+1}+2^{2x+6}=2^{264}\)
\(\Rightarrow2^{2x+1}+2^{2x+1+5}=264\)
\(\Rightarrow2^{2x+1}+2^{2x+1}\cdot2^5=264\)
\(\Rightarrow2^{2x+1}\cdot\left(1+2^5\right)=264\)
\(\Rightarrow2^{2x+1}\cdot26=264\)
\(\Rightarrow2^{2x+1}=\frac{132}{13}\)
\(\Rightarrow2^{2x+1}\notin N\)
\(\Rightarrow2x+1\notin N\)
\(\Rightarrow x\notin N\)
\(\Rightarrow\) Không có giá trị x thỏa mãn.
Vậy không có giá trị x thỏa mãn.
