Question

Va roggggggvggggggggg​

Answer 1

Explicație pas cu pas:

b)

${9}^{x} - 4 \cdot {3}^{x} + 3 \leqslant 0$

$({3}^{x} - 1)({3}^{x} + 3) \leqslant 0$

${3}^{x} = 1 \iff {3}^{x} = {3}^{0} \implies x = 0 \\ {3}^{x} = 3 \iff {3}^{x} = {3}^{1} \implies x = 1$

$1 \leqslant {3}^{x} \leqslant 3$

$\implies x \in \Big[0 ; 1\Big]$

c)

${6}^{x} - {2}^{x + 1} - {3}^{x + 1} + 6 \leqslant 0$

${6}^{x} - 2 \cdot {2}^{x} - 3 \cdot {3}^{x} + 6 \leqslant 0$

$({2}^{x} - 3)({3}^{x} - 2) \leqslant 0$

${2}^{x} - 3 = 0 \iff {2}^{x} = 3 \implies x = log_{2}3 \\ {3}^{x} - 2 = 0 \iff {3}^{x} = 2 \implies x = log_{3}2$

$\implies x \in \Big[\log_{3}2 ; \log_{2}3\Big]$