Question

Resolva as inequações logaritmicas:


a) Log₂ x + Log₂ (x-1) > 1


b) Log$\frac{1}{3}$ ˣ.1 ≥ Log$\frac{1}{3}$ (4x+1)


c) Log2 (x−3).1 ≥ 3


d) Log$\frac{1}{2}$ (2x² + 4x−5) ≥ −4

Answer 1

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$a) \\ \log_2x+\log_2(x+1)\ \textgreater \ 1 \\ \\ C.E. \\ \\ (I)~x\ \textgreater \ 0 \\ (II)~x+1\ \textgreater \ 0 \\ ~~~~~~~x\ \textgreater \ -1 \\ \\ (III)~~base=2~~conserva ~~sinal~~( \ \textgreater \ ) \\ \\ x(x+1)\ \textgreater \ 2 \\ x^2+x-2\ \textgreater \ 0 \\ \\ \triangle=1+8=9 \\ \\ x= \frac{-1\pm3}{2} = \left \{ {{x'= \frac{-1+3}{2}= \frac{2}{2} =1 } \atop {x"= \frac{-1-3}{2}=- \frac{4}{2}=-2 }} \right.$ 

$(I)----------\circ^0xxxxxxxxxx \\ (II)------\circ^{-1}xxxxxxxxxxxxxx \\ (III)xxxx\circ^{-2}--------\circ^1xxxxx \\ \\ S=I\bigcap II \bigcap III \\ \\ S=\{x \in R/ x\ \textgreater \ 1\}$ 

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$b) \\ \log_ \frac{1}{2} }x \geq \log_{ \frac{1}{2} }(4x+1) \\ \\ C.E. \\ (I)~x\ \textgreater \ 0 \\ (II)4x+1\ \textgreater \ 0 \\ ~~~~~4x\ \textgreater \ -1 \\ ~~~~~x\ \textgreater \ - \frac{1}{4} \\ \\ (III)~~base= \frac{1}{2} ~~~troca~~sinal~~de \geq ~~para~~ \leq \\ \\ x \leq 4x+1 \\ x-4x \leq 1 \\ -3x \leq 1~~~\times(-1) \\ \\ 3x \geq -1 \\ x \geq - \frac{1}{3}$ 

$(I)---------\circ^0xxxxxxx \\ (II)-----\circ^{- \frac{1}{4} }xxxxxxxxxxxx \\ (III)--\bullet^{- \frac{1}{3} }xxxxxxxxxxxxxxxx \\ \\ S=I\bigcap II \bigcap III \\ \\ S=\{x\in R /x\ \textgreater \ 0\}$ 
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$c) \\ \log_2(x-3) \geq 3 \\ \\ C.E. \\ (I)x-3\ \textgreater \ 0 \\ ~~~~x\ \textgreater \ 3 \\ \\ (III)~~base=2~~conserva~~sinal~~ \geq \\ \\ x-3 \geq 2^3 \\ x-3 \geq 8 \\ x \geq 8+3 \\ x \geq 11$ 

$(I)---\circ^3xxxxxxxxxxxxxxx \\ (II)-------\bullet^{11}xxxxxx \\ \\ S=I\bigcap II \\ \\ S=\{x\in R /x \geq 11\}$