\(5\cdot3^{x+2}-5=2^4\cdot5^2\)
\(\Rightarrow5\left(3^{x+2}-1\right)=2^4\cdot5^2\)
\(\Rightarrow3^{x+2}-1=\dfrac{2^4\cdot5^2}{5}\)
\(\Rightarrow3^{x+2}-1=16\cdot5\)
\(\Rightarrow3^{x+2}-1=80\)
\(\Rightarrow3^{x+2}=80+1\)
\(\Rightarrow3^{x+2}=81\)
\(\Rightarrow3^{x+2}=3^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
Vậy: x = 2
\(5\cdot3^{x+2}-5=2^4\cdot5^2\\\Rightarrow 5\cdot(3^{x+2}-1)=5\cdot(2^4\cdot5)\\\Rightarrow3^{x+2}-1=2^4\cdot5\\\Rightarrow3^{x+2}-1=16\cdot5\\\Rightarrow3^{x+2}-1=80\\\Rightarrow3^{x+2}=80+1\\\Rightarrow3^{x+2}=81\\\Rightarrow3^{x+2}=3^4\\\Rightarrow x+2=4\\\Rightarrow x=4-2\\\Rightarrow x=2\\Vậy:x=2\)
