Question

07) Determine em R o conjunto solução da inequação log5 (4x − 1) + log5 (x − 5) < 1 a) s = {x ∈ R | 5 < x < 21 4 } b) s = {x ∈ R | 7 < x < 10} c) s = {x ∈ R | 3 < x < 21 4 } d) s = {x ∈ R | − 1 < x < 2} e) s = {x ∈ R | 0 < x < 5}

Answer 1

$\displaystyle \sf \log_5(4x-1)+\log_5(x-5) < 1\\\\ \underline{\text{condi{\c c}{\~a}o de exist{\^e}ncia }} : \\\\ \left \begin{array}{I }\displaystyle \sf x-5>0 \to\boxed{\sf x > 5} \\\\ \displaystyle \sf 4x-1 > 0 \to \boxed{\sf x>\frac{1}{4}}\end{array} \right\} \to \boxed{\sf x > 5} \\\\\\ \log_5(4x-1)+\log_5(x-5) < 1 \\\\ \log_5(4x-1)(x-5) < 1 \\\\ (4x-1)(x-5) < 5^1\\\\ 4x^2-20x-x+5< 5 \\\\ 4x^2-21x < 0 \\\\ x(4x-21)< 0 \\\\ ra{\'i}}zes : \\\\ x = 0 \ \ ;\ \ 4x-21 = 0 \to x=\frac{21}{4}$

$\sf 4x^2-21x < 0$

parábola com concavidade para cima, logo entre as raízes o sinal é negativo ou seja :

$\displaystyle \sf 0<x < \frac{21}{4}$

fazendo intersecção com a condição de existência :

$\displaystyle \sf \left(0<x<\frac{21}{4}\right) \ \cap \ (x > 5 ) = 5<x< \frac{21}{4} \\\\\\\ Portanto : \\\\ \boxed{\sf S=\left\{x\in\mathbb{R}\ |\ 5<x<\frac{21}{4}\right\} }\checkmark$

letra a