Resolva as inequações, no conjunto dos números reais..
Gente por favor quem puder me ajudar, tem haver com propiedades de potência..
Anexos:
![](https://pt-static.z-dn.net/files/dc0/989c6fdfe3b980edd24cd5cba38ff59e.jpg)
![](https://pt-static.z-dn.net/files/da6/713a9d6d4691361ededc552c5899fd29.jpg)
diegobatista123:
(1/8)^2x-5<=(1/4)^x+1. (2^-3)^2x-5<=(2^-2)^x+1. (2)^(-3)×2x-5<=(2)^(-2)×X+1
Soluções para a tarefa
Respondido por
11
E aí mano,
dadas as desigualdades é bom lembrar que quando a base de uma inequação exponencial está compreendida entre 0 e 1, devemos inverter o seu sinal. Vamos relembrar algumas das propriedades da exponenciação, pois vamos utilizar agora:
_______________________
![a^m*a^n~\to~a^{m+n}\\\\
\dfrac{1}{a}~\to~ \dfrac{1}{a^1}~\to~a^{-1}\\\\\\
(a^m)^n~\to~a^{m*n}~\to~a^{mn} a^m*a^n~\to~a^{m+n}\\\\
\dfrac{1}{a}~\to~ \dfrac{1}{a^1}~\to~a^{-1}\\\\\\
(a^m)^n~\to~a^{m*n}~\to~a^{mn}](https://tex.z-dn.net/?f=a%5Em%2Aa%5En%7E%5Cto%7Ea%5E%7Bm%2Bn%7D%5C%5C%5C%5C%0A+%5Cdfrac%7B1%7D%7Ba%7D%7E%5Cto%7E+%5Cdfrac%7B1%7D%7Ba%5E1%7D%7E%5Cto%7Ea%5E%7B-1%7D%5C%5C%5C%5C%5C%5C%0A%28a%5Em%29%5En%7E%5Cto%7Ea%5E%7Bm%2An%7D%7E%5Cto%7Ea%5E%7Bmn%7D++)
_______________________
![\left( \dfrac{1}{8}\right)^{2x-5} \leq ~~\left( \dfrac{1}{4}\right)^{x+1}\\\\\\
\left( \dfrac{1}{2^3}\right)^{2x-5} \leq ~~\left( \dfrac{1}{2^2}\right)^{x+1}\\\\\\
(2^{-3})^{2x-5} \geq ~~(2^{-2})^{x+1}\\\\
\not2^{-6x+15} \geq ~~\not2^{-2x-2}\\\\
-6x+15 \geq -2x-2\\
15 +2\geq-2x+6x\\
17 \geq 4x\\\\
x \geq \dfrac{17}{4}\\\\\\
\boxed{S=\left\{x\in~\mathbb{R}~|~x \geq \dfrac{17}{4}\right\}}
\left( \dfrac{1}{8}\right)^{2x-5} \leq ~~\left( \dfrac{1}{4}\right)^{x+1}\\\\\\
\left( \dfrac{1}{2^3}\right)^{2x-5} \leq ~~\left( \dfrac{1}{2^2}\right)^{x+1}\\\\\\
(2^{-3})^{2x-5} \geq ~~(2^{-2})^{x+1}\\\\
\not2^{-6x+15} \geq ~~\not2^{-2x-2}\\\\
-6x+15 \geq -2x-2\\
15 +2\geq-2x+6x\\
17 \geq 4x\\\\
x \geq \dfrac{17}{4}\\\\\\
\boxed{S=\left\{x\in~\mathbb{R}~|~x \geq \dfrac{17}{4}\right\}}](https://tex.z-dn.net/?f=%5Cleft%28+%5Cdfrac%7B1%7D%7B8%7D%5Cright%29%5E%7B2x-5%7D+%5Cleq+%7E%7E%5Cleft%28+%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%5E%7Bx%2B1%7D%5C%5C%5C%5C%5C%5C%0A%5Cleft%28+%5Cdfrac%7B1%7D%7B2%5E3%7D%5Cright%29%5E%7B2x-5%7D+%5Cleq+%7E%7E%5Cleft%28+%5Cdfrac%7B1%7D%7B2%5E2%7D%5Cright%29%5E%7Bx%2B1%7D%5C%5C%5C%5C%5C%5C%0A%282%5E%7B-3%7D%29%5E%7B2x-5%7D++%5Cgeq++%7E%7E%282%5E%7B-2%7D%29%5E%7Bx%2B1%7D%5C%5C%5C%5C%0A%5Cnot2%5E%7B-6x%2B15%7D+%5Cgeq+%7E%7E%5Cnot2%5E%7B-2x-2%7D%5C%5C%5C%5C%0A-6x%2B15+%5Cgeq+-2x-2%5C%5C%0A15+%2B2%5Cgeq-2x%2B6x%5C%5C%0A17+%5Cgeq+4x%5C%5C%5C%5C%0Ax+%5Cgeq++%5Cdfrac%7B17%7D%7B4%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5Cleft%5C%7Bx%5Cin%7E%5Cmathbb%7BR%7D%7E%7C%7Ex+%5Cgeq++%5Cdfrac%7B17%7D%7B4%7D%5Cright%5C%7D%7D+++%0A++++++)
_______________________
![( \sqrt{3})^{ x^{2} -3x} \geq ( \sqrt{3})^{4}~~~(bases~iguais~eliminamos)\\\\
x^{2} -3x \geq 4\\
x^{2} -3x-4 \geq 0\\\\
\Delta=(-3)^2-4*1*(-4)\\
\Delta=9+16\\
\Delta=25\\\\
x \geq \dfrac{-(-3)\pm \sqrt{25} }{2*1} \geq \dfrac{3\pm5}{2} \geq \begin{cases}x' \geq \dfrac{3-5}{2} \geq \dfrac{-2}{~~2} \geq -1\\\\
x'' \geq \dfrac{3+5}{2} \geq \dfrac{8}{2} \geq 4 \end{cases}\\\\\\
\boxed{S=\left\{x\in~\mathbb{R}~|~4 \leq x \geq -1\right\}} ( \sqrt{3})^{ x^{2} -3x} \geq ( \sqrt{3})^{4}~~~(bases~iguais~eliminamos)\\\\
x^{2} -3x \geq 4\\
x^{2} -3x-4 \geq 0\\\\
\Delta=(-3)^2-4*1*(-4)\\
\Delta=9+16\\
\Delta=25\\\\
x \geq \dfrac{-(-3)\pm \sqrt{25} }{2*1} \geq \dfrac{3\pm5}{2} \geq \begin{cases}x' \geq \dfrac{3-5}{2} \geq \dfrac{-2}{~~2} \geq -1\\\\
x'' \geq \dfrac{3+5}{2} \geq \dfrac{8}{2} \geq 4 \end{cases}\\\\\\
\boxed{S=\left\{x\in~\mathbb{R}~|~4 \leq x \geq -1\right\}}](https://tex.z-dn.net/?f=%28+%5Csqrt%7B3%7D%29%5E%7B+x%5E%7B2%7D+-3x%7D+%5Cgeq+%28+%5Csqrt%7B3%7D%29%5E%7B4%7D%7E%7E%7E%28bases%7Eiguais%7Eeliminamos%29%5C%5C%5C%5C%0A+x%5E%7B2%7D+-3x+%5Cgeq+4%5C%5C%0A+x%5E%7B2%7D+-3x-4+%5Cgeq+0%5C%5C%5C%5C%0A%5CDelta%3D%28-3%29%5E2-4%2A1%2A%28-4%29%5C%5C%0A%5CDelta%3D9%2B16%5C%5C%0A%5CDelta%3D25%5C%5C%5C%5C%0Ax+%5Cgeq++%5Cdfrac%7B-%28-3%29%5Cpm+%5Csqrt%7B25%7D+%7D%7B2%2A1%7D+%5Cgeq++%5Cdfrac%7B3%5Cpm5%7D%7B2%7D++%5Cgeq+%5Cbegin%7Bcases%7Dx%27+%5Cgeq++%5Cdfrac%7B3-5%7D%7B2%7D+%5Cgeq++%5Cdfrac%7B-2%7D%7B%7E%7E2%7D+%5Cgeq+-1%5C%5C%5C%5C%0Ax%27%27+%5Cgeq++%5Cdfrac%7B3%2B5%7D%7B2%7D+%5Cgeq++%5Cdfrac%7B8%7D%7B2%7D+%5Cgeq+4++++%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5Cleft%5C%7Bx%5Cin%7E%5Cmathbb%7BR%7D%7E%7C%7E4+%5Cleq+x+%5Cgeq+-1%5Cright%5C%7D%7D+++)
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![5^{2x}<-5+6*5^x\\
(5^x)^2-6*5^x+5<0\\\\
5^x=k,~substitua:\\\\
k^2-6k+5<0\\\\
\Delta=(-6)^2-4*1*5\\
\Delta=36-20\\
\Delta=16\\\\
k< \dfrac{-(-6)\pm \sqrt{16} }{2*1}< \dfrac{6\pm4}{2}<\begin{cases} k'<\dfrac{6-4}{2}< \dfrac{2}{2}<1\\\\
k''< \dfrac{6+4}{2}< \dfrac{10}{2}<5 \end{cases} 5^{2x}<-5+6*5^x\\
(5^x)^2-6*5^x+5<0\\\\
5^x=k,~substitua:\\\\
k^2-6k+5<0\\\\
\Delta=(-6)^2-4*1*5\\
\Delta=36-20\\
\Delta=16\\\\
k< \dfrac{-(-6)\pm \sqrt{16} }{2*1}< \dfrac{6\pm4}{2}<\begin{cases} k'<\dfrac{6-4}{2}< \dfrac{2}{2}<1\\\\
k''< \dfrac{6+4}{2}< \dfrac{10}{2}<5 \end{cases}](https://tex.z-dn.net/?f=5%5E%7B2x%7D%26lt%3B-5%2B6%2A5%5Ex%5C%5C%0A%285%5Ex%29%5E2-6%2A5%5Ex%2B5%26lt%3B0%5C%5C%5C%5C%0A5%5Ex%3Dk%2C%7Esubstitua%3A%5C%5C%5C%5C%0Ak%5E2-6k%2B5%26lt%3B0%5C%5C%5C%5C%0A%5CDelta%3D%28-6%29%5E2-4%2A1%2A5%5C%5C%0A%5CDelta%3D36-20%5C%5C%0A%5CDelta%3D16%5C%5C%5C%5C%0Ak%26lt%3B+%5Cdfrac%7B-%28-6%29%5Cpm+%5Csqrt%7B16%7D+%7D%7B2%2A1%7D%26lt%3B+%5Cdfrac%7B6%5Cpm4%7D%7B2%7D%26lt%3B%5Cbegin%7Bcases%7D+k%27%26lt%3B%5Cdfrac%7B6-4%7D%7B2%7D%26lt%3B+%5Cdfrac%7B2%7D%7B2%7D%26lt%3B1%5C%5C%5C%5C%0Ak%27%27%26lt%3B+%5Cdfrac%7B6%2B4%7D%7B2%7D%26lt%3B+%5Cdfrac%7B10%7D%7B2%7D%26lt%3B5++++%5Cend%7Bcases%7D++)
Retomando a variável original, teremos:
![5^x<1~~~~~~~~~~~~~~~~~~~~~5^x<5\\
\not5^x<\not5^0~~~~~~~~~~~~~~~~~~\not5^x<\not5^1\\\\
x<0~~~~~~~~~~~~~~~~~~~~~~~~x<1\\\\\\
\boxed{S=\{x\in~\mathbb{R}~|~0>x<1\}} 5^x<1~~~~~~~~~~~~~~~~~~~~~5^x<5\\
\not5^x<\not5^0~~~~~~~~~~~~~~~~~~\not5^x<\not5^1\\\\
x<0~~~~~~~~~~~~~~~~~~~~~~~~x<1\\\\\\
\boxed{S=\{x\in~\mathbb{R}~|~0>x<1\}}](https://tex.z-dn.net/?f=5%5Ex%26lt%3B1%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E5%5Ex%26lt%3B5%5C%5C%0A%5Cnot5%5Ex%26lt%3B%5Cnot5%5E0%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5Cnot5%5Ex%26lt%3B%5Cnot5%5E1%5C%5C%5C%5C%0Ax%26lt%3B0%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7Ex%26lt%3B1%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5C%7Bx%5Cin%7E%5Cmathbb%7BR%7D%7E%7C%7E0%26gt%3Bx%26lt%3B1%5C%7D%7D)
Tenha ótimos estudos =))
dadas as desigualdades é bom lembrar que quando a base de uma inequação exponencial está compreendida entre 0 e 1, devemos inverter o seu sinal. Vamos relembrar algumas das propriedades da exponenciação, pois vamos utilizar agora:
_______________________
_______________________
_______________________
_______________________
Retomando a variável original, teremos:
Tenha ótimos estudos =))
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