Question

3. (ITA) - resolva a inequação em R:16<(41​)log1/5​(x2−x+19)​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf{16 < \left(\dfrac{1}{4}\right)^{log_{\frac{1}{5}}\:(x^2 - x + 19)}$

$\sf{16 < \left(\dfrac{1}{4}\right)^{-log_{5}\:(x^2 - x + 19)}$

$\sf{\left(\dfrac{1}{4}\right)^{-log_{5}\:(x^2 - x + 19)} > 16}$

$\sf{4^{log_{5}\:(x^2 - x + 19)} > 4^2}$

$\sf{log_{5}\:(x^2 - x + 19) > 2}$

$\sf{log_{5}\:(x^2 - x + 19) > log_5\:5^2}$

$\sf{x^2 - x + 19 > 25}$

$\sf{x^2 - x - 6 > 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-1)^2 - 4.1.(-6)}$

$\mathsf{\Delta = 1 + 24}$

$\mathsf{\Delta = 25}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{25}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{1 + 5}{2} = \dfrac{6}{2} = 3}\\\\\mathsf{x'' = \dfrac{1 - 5}{2} = -\dfrac{4}{2} = -2}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{\:\:\:]-\infty,-2\:[ \:\:\:\cup\:\:\: ]\:3,+\infty[\:\:\:\}}}}$