Question

16^x+5*4^(x+1)- 21=0

Answer 1

$\it 16^x+5\cdot4^{x+1}-21=0 \Rightarrow (4^2)^x+5\cdot4\cdot4^x-21=0 \Rightarrow (4^x)^2+20\cdot4^x-21=0\\ \\ Not\breve am\ \ 4^x=t,\ \ t>0,\ \ iar\ ecua\c{\it t}ia\ devine:\\ \\ t^2+20t-21=0 \Rightarrow t^2+21t-t-21=0 \Rightarrow t(t+21)-(t+21)=0 \Rightarrow \\ \\ \Rightarrow (t+21)(t-1)=0 \Rightarrow \begin{cases}\it t+21=0 \Rightarrow t=-21<0\ nu\ convine\\ \\ t-1=0 \Rightarrow t=1\end{cases}$

$\it t=1 \Rightarrow 4^x=1 \Rightarrow x=0$