[tex] log_{4}(3^x -1)* log_{4} \frac{16}{3^x-1} \leq \frac{3}{4} \\ log_{4}(3^x-1)( log_{4}16- log_{4}(3^x-1)) \leq \frac{3}{4} \\ 2 log_{4}(3^x-1)- log_{4} ^{2}(3^x-1) \leq \frac{3}{4} \\ fie: log_{4}(3^x-1)=t \\ -t^2+2t- \frac{3}{4} \leq 0 \\ 4t^2-8t+3 \geq 0 \\ [/tex]
t∈(-∞;1/2]∪[3/2;∞)
revenim la substitutie
[tex] log_{4}(3^x-1) \leq \frac{1}{2} sau log_{4}(3^x-1) \geq \frac{3}{2} \\ 3^x-1 \leq 2sau3^x-1 \geq 8 \\ 3^x \leq 3sau3^x \geq 9 \\ x \leq 1saux \geq 2[/tex]
raspuns:x∈(-∞;1]∪[2;∞)
