Question

Sa se rezolve: 4 ^ (x+1) -5*2^x+1=0

Answer 1

[tex]4^{x+1}-5\cdot 2^{x}+1=0\\ 4^x\cdot 4-5\cdot 2^x+1=0\\ 4\cdot (2^x)^2-5\cdot 2^x+1=0\\ \text{Notam}\ 2^x=t\ \textgreater \ 0\\ 4t^2-5t+1=0\\ \Delta=25-16=9\Rightarrow \sqrt{\Delta}=3\\ t_1=\dfrac{5+3}{8}=1\\ t_2=\dfrac{5-3}{8}=\dfrac{1}{4}\\ 2^x=1\Rightarrow x=0\\ 2^x=\dfrac{1}{4}\Rightarrow x=-2[/tex]