Question

Mă poate ajuta cineva?

Answer 1

Răspuns:

Explicație pas cu pas:

$\log_2(4^x-1)\cdot\log_2(4^{x+1}-4)\geq -1\\\texttt{Conditii de existenta:}x>0\\\log_2(4^x-1)\cdot\log_2{[4(4^x-1)]}\geq -1\\\log_2(4^x-1)\cdot [\log_24+\log_2(4^x-1)]}\geq -1\\\log_2(4^x-1)\cdot{[2+\log_2(4^x-1)]}\geq -1\\\log_2(4^x-1)\stackrel{not}{=}t\\\texttt{Inecuatia devine :}\\t(2+t)\geq -1\\t^2+2t+1\geq 0\\(t+1)^2\geq 0,\texttt{Relatia este adevarata pentru orice t numar real}\\\texttt{Prin urmare }x\in(0,\infty),\texttt{ deci raspunsul este b)}$

Answer 2

$\log_{2}(4^x-1)\cdot \log_2{(4^{x+1}-4)}\geq -1\\ \\ \log_{2}(4^x-1)\cdot \log_2{(4\cdot 4^{x}-4)}\geq -1 \\ \\ \log_{2}(4^x-1)\cdot \log_2\Big[{4\cdot (4^{x}-1)\Big]}\geq -1 \\ \\ \log_{2}(4^x-1)\Big[\log_2 4+\log_{2}(4^{x}-1)\Big]\geq -1\\ \\ \log_{2}(4^x-1) = t \\ \\ t(2+t)\geq -1 \\ t^2+2t+1 \geq 0 \\ (t+1)^2\geq 0\\ \\ \Rightarrow t\in \mathbb{R} \Rightarrow \log_{2}(4^x-1) \in \mathbb{R}\\ \\ \text{Dar conditia de existenta e ca }4^x-1 > 0 \Rightarrow 4^x > 1 \Rightarrow \\ \\ \Rightarrow \boxed{x > 0}$

$\Rightarrow b)\,\text{corect}$