Question

rezolvati in multimea numerelor reale ecuatiia 4^x - 2^(x+1) -3=0

Answer 1

4^x - 2^(x+1) - 3 = 0

(2^2)^x - 2^x * 2 - 3 =0

(2^x)^2 - 2^x *2 - 3 = 0 notezi pe t = 2^x

t^2 - 2t - 3 = 0

delta = 4+12 = 16

t1 = (2+4)/2 = 6/2 = 3

t2 = (2-4)/2 = -2/2 = -1

Revin la substitutie

2^x = -1, x nu apartine R

2^x = 3, x = log₂ 3

S={log₂ 3}

Answer 2

$\it 4^x-2^{x+1}-3=0 \Rightarrow (2^2)^x-2^x\cdot2^1-3=0 \Rightarrow (2^x)^2-2\cdot2^x-3=0\\ \\ Not\breve am\ 2^x=t,\ t>0,\ iar\ ecua \c{\it t}ia\ devine:\\ \\ t^2-2t-3=0 \Rightarrow t^2+t-3t-3=0 \Rightarrow t(t+1)-3(t+1)=0 \Rightarrow \\ \\ \Rightarrow (t+1)(t-3)=0 \Rightarrow \begin{cases} \it t+1=0 \Rightarrow t=-1<0,\ nu\ convine\\ \\ \it t-3=0 \Rightarrow t=3 \Rightarrow 2^x=3 \Rightarrow x=log_2 3\end{cases}$