Question

Resolva as inequações:


a) Log₂ x + Log₂ (x-1) > 1


b) Log1/3 (x).1 ≥ Log1/3 (4x+1)


c) Log2 (x−3).1 ≥ 3


d) Log1/2 (2x(2) + 4x−5) ≥ −4

Answer 1

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

$(I)-----\circ^0 xxxxxxxxxxxxx \\ (II)--------\circ^1xxxxxxx \\ (III)--\circ^{-1}xxxxxxxxxxx\circ^2-- \\ \\ S={\{x\in R /1\ \textless \ x\ \textless \ 2\}$ 
(I)∩(II)∩(III)=1<x<2
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$b) \\ C.E. \\ (I)~~x\ \textgreater \ 0 \\ (II)4x+1=0 \\ ~~~~~~ 4x=-1 \\ ~~~~~~x=- \frac{1}{4} \\ \\ \log_{ \frac{1}{3}x \geq \log_{ \frac{1}{3}}(4x+1)} \\ \\ base= \frac{1}{3} ~~decrescente~~troca~~sinal \\ \\(III)~~ x \leq 4x+1 \\ \\ x-4x \leq 1 \\ \\ -3x \leq 1~~\div(-1) \\ \\ 3x \geq -1 \\ \\ x \geq - \frac{1}{3} \\ \\ (I)----------\circ^0xxxxx \\ (II)------\circ^{- \frac{1}{4}} xxxxxxxxxx \\ \\ (III)---\bullet^{- \frac{1}{3} }xxxxxxxxxxxxxxx \\ \\ S=\{x\in R/x\ \textgreater \ 0\}$

(I)∩(II)∩(III)=x>0

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$c) \\ C.E. \\ (I)~~ x-3\ \textgreater \ 0 \\ ~~~~~~~x\ \textgreater \ 3 \\ \\(II)~~ \log_2(x-3) \geq 3 \\ \\ base=2~~crescente \\ \\ x-3 \geq 2^3 \\ \\ x-3 \geq 8 \\ \\ x \geq 8+3 \\ \\ x \geq 11 \\ \\ (I)~~--\circ^3xxxxxxxx \\ (II)----\bullet^{11}xxxx \\ \\ S=\{x\in R/x \geq 11\}$