Răspuns :
[tex] Nicio~problema~:\\ \\ 4^x+2^{x+1}=80\\ \\ 2^{2x}+2*2^x=80.\\ \\ Notezi~astfel:~2^x=t,~t\in~N^{*}.\\ \\ t^2+2t-80=0\\ \\ \Delta=4+320=324=18^2.\\ \\ t_1=\frac{-2+18}{2}= 8\\ \\ t_2=\frac{-2-18}{2}=-10.\\ \\ Deoarece~2^x>0~\Leftrightarrow~t>0,~t_2~nu~convine!\\ \\ Solutia~unica~este~t_1=8~si~din~egalitatea~2^x=8\Rightarrow~x=3. [/tex]
4^x + 2^(x+1) = 80
(2^2)^x + 2^x ×2 = 80
2^x = a
a² + 2a = 80
a² + 2a - 80 = 0
Δ = b² - 4ac = 4 + 320 = 324
x1 = (-2 + 18)/2 = 16/2 = 8
x2 = (-2 - 18)/2 = -20 / 2 = -10 (nu este solutie )
2^x = 8
2^x = 2^3
x = 3